Solutions_Manual_Principles_and_Modern_A

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1.1a. Concentration of a gas mi[ture.
A Pi[WXUe Rf QRbOe gaVeV [heOiXP (1), aUgRQ (2), NU\SWRQ (3), aQd [eQRQ (4)] iV aW a WRWaO SUeVVXUe Rf
200 NPa aQd a WePSeUaWXUe Rf 400 K. If Whe Pi[WXUe haV eTXaO PROe fUacWiRQV Rf each Rf Whe gaVeV,
deWeUPiQe:
a) The cRPSRViWiRQ Rf Whe Pi[WXUe iQ WeUPV Rf PaVV fUacWiRQV.
SROXWiRQ: 
BaViV: 100 NPROe Rf Whe Pi[WXUe
b) The aYeUage PROecXOaU ZeighW Rf Whe Pi[WXUe.
SROXWiRQ
c) The WRWaO PROaU cRQceQWUaWiRQ.
SROXWiRQ
d) The PaVV deQViW\.
SROXWiRQ
1.2a. ConcenWUaWion of a liTXid VolXWion fed Wo a diVWillaWion colXmn.
A VolXWion of caUbon WeWUachloUide (1) and caUbon diVXlfide (2) conWaining 50% b\ ZeighW each iV Wo
be conWinXoXVl\ diVWilled aW Whe UaWe of 4,000 kg/h DeWeUmine:
a) The concenWUaWion of Whe mi[WXUe in WeUmV of mole fUacWionV.
SolXWion
BaViV: 100 kg mi[WXUe
b) The aYeUage molecXlaU ZeighW of Whe mi[WXUe.
SolXWion
c) CalcXlaWe Whe feed UaWe in kmol/h.
SolXWion
1.3a. ConcenWUaWion of liTXified naWXUal gaV.
A VaPSOe Rf OiTXified QaWXUaO gaV, LNG, fURP AOaVNa haV Whe fROORZiQg PROaU cRPSRViWiRQ: 93.5%
CH4, 4.6% C2H6, 1.2% C3H8, aQd 0.7% CO2. CaOcXOaWe:
a) AYeUage PROecXOaU ZeighW Rf Whe LNG Pi[WXUe.
SROXWiRQ
b) WeighW fUacWiRQ Rf CH4 iQ Whe Pi[WXUe.
SROXWiRQ
BaViV: 100 NPROeV Rf LNG
c) The LNG iV heaWed WR 300 K aQd 140 NPa, aQd YaSRUi]eV cRPSOeWeO\. EVWiPaWe Whe deQViW\ Rf Whe
gaV Pi[WXUe XQdeU WheVe cRQdiWiRQV.
SROXWiRQ
1.4b. ConcenWUaWion of a flXe gaV.
A flXe gaV conViVWV of caUbon dio[ide, o[\gen, ZaWeU YapoU, and niWUogen. The molaU fUacWionV of CO2
and O2 in a Vample of Whe gaV aUe 12% and 6%, UeVpecWiYel\. The ZeighW fUacWion of H2O in Whe gaV
iV 6.17%. EVWimaWe Whe denViW\ of WhiV gaV aW 500 K and 110 kPa.
SolXWion
BaViV: 100 kmole of gaV mi[WXUe
LeW [ = molaU fUacWion of ZaWeU in Whe mi[WXUe (aV a peUcenW)
IniWial eVWimaWe
FUom Whe giYen ZaWeU ZeighW fUacWion (0.0617):
1.5b. Material balances around an ammonia gas absorber.
A gaV VWUeaP fORZV aW Whe UaWe Rf 10.0 P3/V aW 300 K aQd 102 NPa. IW cRQViVWV Rf aQ eTXiPROaU
Pi[WXUe Rf aPPRQia aQd aiU. The gaV eQWeUV WhURXgh Whe bRWWRP Rf a SacNed bed gaV abVRUbeU
ZheUe iW fORZV cRXQWeUcRXUUeQW WR a VWUeaP Rf SXUe OiTXid ZaWeU WhaW abVRUbV 90% Rf aOO Rf Whe
aPPRQia, aQd YiUWXaOO\ QR aiU. The abVRUbeU YeVVeO iV c\OiQdUicaO ZiWh aQ iQWeUQaO diaPeWeU Rf 2.5 P.
a) NegOecWiQg Whe eYaSRUaWiRQ Rf ZaWeU, caOcXOaWe Whe aPPRQia PRO fUacWiRQ iQ Whe gaV OeaYiQg Whe
abVRUbeU.
SROXWiRQ:
BaViV: 1 VecRQd
A = aPPRQia B = aiU
b) CaOcXOaWe Whe RXWOeW gaV PaVV YeORciW\ (defiQed aV PaVV fORZ UaWe SeU XQiW ePSW\ WXbe
cURVV-VecWiRQaO aUea).
SROXWiRQ:
1.6b. Velocities and flu[es in a gas mi[ture.
A gaV Pi[WXUe aW a WRWaO SUeVVXUe Rf 150 NPa aQd 295 K cRQWaiQV 20% H2, 40% O2, aQd 40% H2O
b\ YROXPe. The abVROXWe YeORciWieV Rf each VSecieV aUe -10 P/V, -2 P/V, aQd 12 P/V, UeVSecWiYeO\, aOO
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. PUopeUWieV of aiU VaWXUaWed ZiWh ZaWeU YapoU.
AiU, VWRUed iQ a 30-P3 cRQWaiQeU aW 340 K aQd 101.3 NPa iV VaWXUaWedZiWh ZaWeU YaSRU. DeWeUPiQe Whe
fROORZiQg SURSeUWieV Rf Whe gaV Pi[WXUe:
a) MROe fUacWiRQ Rf ZaWeU YaSRU.
b) AYeUage PROecXOaU ZeighW Rf Whe Pi[WXUe.
c) TRWaO PaVV cRQWaiQed iQ Whe WaQN.
d) MaVV Rf ZaWeU YaSRU iQ Whe WaQN.
SROXWiRQ
a) AQWRiQe eTXaWiRQ fRU ZaWeU YaSRU:
FRU a VaWXUaWed Pi[WXUe
b)
c)
d)
1.8c. WaWeU balance aUoXnd an indXVWUial cooling WoZeU.
The cRROiQg ZaWeU fORZ UaWe WR Whe cRQdeQVeUV Rf a big cRaO-fiUed SRZeU SOaQW iV 8,970 Ng/V. The
ZaWeU eQWeUV Whe cRQdeQVeUV aW 29 qC aQd OeaYeV aW 45 qC. FURP Whe cRQdeQVeUV, Whe ZaWeU fORZV WR
a cRROiQg WRZeU ZheUe iW iV cRROed dRZQ bacN WR 29 qC b\ cRXQWeUcRXUUeQW cRQWacW ZiWh aiU (Vee
FigXUe 1.11). The aiU eQWeUV Whe cRROiQg WRZeU aW Whe UaWe Rf 6,500 Ng/V Rf dU\ aiU, aW a dU\-bXOb
WePSeUaWXUe Rf 30 qC aQd a hXPidiW\ Rf 0.016 Ng Rf ZaWeU/Ng Rf dU\ aiU. IW OeaYeV Whe cRROiQg WRZeU
VaWXUaWed ZiWh ZaWeU YaSRU aW 38 qC.
a) CaOcXOaWe Whe ZaWeU ORVVeV b\ eYaSRUaWiRQ iQ Whe cRROiQg WRZeU.
SROXWiRQ
CRQVideU Whe aiU OeaYiQg Whe WRZeU VaWXUaWed aW 38 qC.
AQWRiQe eTXaWiRQ fRU ZaWeU YaSRU:
E = ZaWeU ORVW b\ eYaSRUaWiRQ iQ Whe aiU
b) TR accRXQW fRU ZaWeU ORVVeV iQ Whe cRROiQg WRZeU, SaUW Rf Whe effOXeQW fURP a QeaUb\ PXQiciSaO
ZaVWeZaWeU WUeaWPeQW SOaQW ZiOO be XVed aV PaNeXS ZaWeU. ThiV PaNeXS ZaWeU cRQWaiQV 500 Pg/L
Rf diVVROYed VROidV. TR aYRid fRXOiQg Rf Whe cRQdeQVeU heaW-WUaQVfeU VXUfaceV, Whe ciUcXOaWiQg ZaWeU
iV WR cRQWaiQ QR PRUe WhaQ 2,000 Pg/L Rf diVVROYed VROidV. TheUefRUe, a VPaOO aPRXQW Rf Whe
ciUcXOaWiQg ZaWeU PXVW be deOibeUaWeO\ diVcaUded (bORZdRZQ). WiQdage ORVVeV fURP Whe WRZeU aUe
eVWiPaWed aW 0.2% Rf Whe UeciUcXOaWiRQ UaWe. EVWiPaWe Whe PaNeXS-ZaWeU UeTXiUePeQW.
SROXWiRQ
W = ZiQdage ORVVeV [P = 500 SSP [c = 2000 SSP
M = PaNeXS ZaWeU UaWe B = bORZdRZQ UaWe
IQiWiaO eVWiPaWeV
WaWeU baOaQce
SROidV baOaQce:
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
CalcXlaWe ZaWeU YaSRU SUeVVXUe aW 350 K
AQWRiQe eTXaWiRQ fRU ZaWeU YaSRU:
1.10b. ActiYated carbon adsorption; material balances.
A ZaVWe gaV cRQWaiQV 0.3% WROXeQe iQ aiU, aQd RccXSieV a YROXPe Rf 2,500 P3 aW 298 K aQd 101.3
NPa. IQ aQ effRUW WR UedXce Whe WROXeQe cRQWeQW Rf WhiV gaV, iW iV e[SRVed WR 100 Ng Rf acWiYaWed
caUbRQ, iQiWiaOO\ fUee Rf WROXeQe. The V\VWeP iV aOORZed WR Ueach eTXiOibUiXP aW cRQVWaQW WePSeUaWXUe
aQd SUeVVXUe. AVVXPiQg WhaW Whe aiU dReV QRW adVRUb RQ Whe caUbRQ, caOcXOaWe Whe eTXiOibUiXP
cRQceQWUaWiRQ Rf WROXeQe iQ Whe gaVeRXV ShaVe, aQd Whe aPRXQW Rf WROXeQe adVRUbed b\ Whe caUbRQ.
The adVRUSWiRQ eTXiOibUiXP fRU WhiV V\VWeP iV giYeQ b\ Whe FUeXQdOich iVRWheUP(EAB CRQWURO CRVW
MaQXaO, 3Ud. ed., U. S. E. P. A., ReVeaUch TUiaQgOe PaUN, NC, 1987.):
ZheUe W iV Whe caUbRQ eTXiOibUiXP adVRUSWiYiW\, iQ Ng Rf WROXeQe/Ng Rf caUbRQ, aQd S* iV Whe
eTXiOibUiXP WROXeQe SaUWiaO SUeVVXUe, iQ Pa, aQd PXVW be beWZeeQ 0.7 aQd 345 Pa.
SROXWiRQ
M = PaVV Rf caUbRQ
[ = PROeV Rf WROXeQe adVRUbed
IQiWiaO eVWiPaWeV
1.11b. ActiYated 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.
Solution
1.12a, d. Estimation of gas diffusivit\ b\ the Wilke-Lee equation.
E. M. LaUVRQ (MS WheViV, OUegRQ SWaWe UQiYeUViW\, 1964) PeaVXUed Whe diffXViYiW\ Rf chORURfRUP
iQ aiU aW 298 K aQd 1 aWP aQd UeSRUWed iWV YaOXe aV 0.093 cP2/V. EVWiPaWe Whe diffXViRQ
cRefficieQW b\ Whe WiONe-Lee eTXaWiRQ aQd cRPSaUe iW ZiWh Whe e[SeUiPeQWaO YaOXe.
SROXWiRQ
FURP
ASSeQdi[
B
1.13a, d. Estimation of gas diffusivit\ b\ the Wilke-Lee equation.
a) EVWimaWe Whe diffXViYiW\ of naphWhalene (C10H8) in aiU aW 303 K and 1 baU. CompaUe iW ZiWh Whe
e[peUimenWal YalXe of 0.087 cm2/V UepoUWed in Appendi[ A. The noUmal boiling poinW of
naphWhalene iV 491.1 K, and iWV cUiWical YolXme iV 413 cm3/mol.
SolXWion
E[peUimenWal YalXe
b) EVWimaWe WhediffXViYiW\ of p\Uidine (C5H5N) in h\dUogen aW 318 K and 1 aWm. CompaUe iW ZiWh Whe
e[peUimenWal YalXe of 0.437 cm2/V UepoUWed in Appendi[ A. The noUmal boiling poinW of p\Uidine iV
388.4 K, and iWV cUiWical YolXme iV 254 cm3/mol.
SolXWion
E[peUimenWal YalXe
c) EVWimaWe Whe diffXViYiW\ of aniline (C6H7N) in aiU aW 273 K and 1 aWm. CompaUe iW ZiWh Whe
e[peUimenWal YalXe of 0.061 cm2/V (GXilliland, E. R., IQd. EQg. CheP., 2�:681, 1934). The
noUmal boiling poinW of aniline iV 457.6 K, and iWV cUiWical YolXme iV 274 cm3/mol.
SolXWion
E[peUimenWal YalXe
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 DeYelop., 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 u 10
-25 (J-
m3)1/2]
a) Modify the MathcadŽ 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)
Solution
b) EVWiPaWe Whe diffXViRQ cRefficieQW fRU a Pi[WXUe Rf PeWh\O chORUide aQd VXOfXU diR[ide aW 1 baU aQd
323 K, aQd cRPSaUe iW WR Whe e[SeUiPeQWaO YaOXe Rf 0.078 cP2/V. The daWa UeTXiUed WR XVe BURNaZ'V
UeOaWiRQ aUe VhRZQ beORZ (Reid, eW aO., 1987):
PaUaPeWeU MeWh\O chORUide SXOfXU diR[ide
Tb , K 249.1 263.2
Vb , cP
3/PRO 50.6 43.8
PS, deb\eV 1.9 1.6
M 50.5 64.06
SROXWiRQ
1.15d. Diffusivit\ of polar gases 
EYaOXaWe Whe diffXViRQ cRefficieQW Rf h\dURgeQ chORUide iQ ZaWeU aW 373 K aQd 1 baU. The daWa UeTXiUed
WR XVe BURNaZ'V UeOaWiRQ (Vee PURbOeP 1.14) aUe VhRZQ beORZ (Reid, eW aO., 1987):
PaUaPeWeU H\dURgeQ chORUide WaWeU
Tb , K 188.1 373.2
Vb , cP3/PRO 30.6 18.9
PS, deb\eV 1.1 1.8
M 36.5 18
SROXWiRQ
1.16d. Diffusivit\ of polar gases 
EYaOXaWe Whe diffXViRQ cRefficieQW Rf h\dURgeQ VXOfide iQ VXOfXU diR[ide aW 298 K aQd 1.5 baU. The daWa
UeTXiUed WR XVe BURNaZ'V UeOaWiRQ (VeePURbOeP 1.14) aUe VhRZQ beORZ (Reid, eW aO., 1987):
PaUaPeWeU H\dURgeQ VXOfide SXOfXU diR[ide
Tb , K 189.6 263.2
Vb , cP3/PRO 35.03 43.8
PS, deb\eV 0.9 1.6
M 34.08 64.06
SROXWiRQ
1.17a,d. Effective diffusivit\ in a multicomponent stagnant gas mi[ture.
Calculate the effective diffusivit\ of nitrogen through a stagnant gas mi[ture at 373 K and 1.5 bar.
The mi[ture composition is:
O2 15 mole %
CO 30%
CO2 35%
N2 20%
Sol;ution
Calculate mole fractions on a nitrogen (1)-free basis:
o[\gen (2); carbon mono[ide (3); carbon dio[ide (4)
Calculate binar\ MS diffusivities from Wilke-Lee equation
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. AiU & WaVWe
Manage. AVVoc., 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
HgCl2 (1) O2 (2) CO2 (3) H2O (4) N2 (5)
1.19a. Wilke-Chang method for liquid diffusivit\.
EVWiPaWe Whe OiTXid diffXViYiW\ Rf caUbRQ WeWUachORUide iQ diOXWe VROXWiRQ iQWR eWhaQRO aW 298 K.
CRPSaUe WR Whe e[SeUiPeQWaO YaOXe UeSRUWed b\ Reid, eW aO. (1987) aV 1.5 u 10-5 cP2/V. The cUiWicaO
YROXPe Rf caUbRQ WeWUachORUide iV 275.9 cP3/PRO. The YiVcRViW\ Rf OiTXid eWhaQRO aW 298 K iV 1.08 cP.
SROXWiRQ
1.20b. Diffusion in electrol\te solutions.
WheQ a VaOW diVVRciaWeV iQ VROXWiRQ, iRQV UaWheU WhaQ PROecXOeV diffXVe. IQ Whe abVeQce Rf aQ
eOecWUic SRWeQWiaO, Whe diffXViRQ Rf a ViQgOe VaOW Pa\ be WUeaWed aV PROecXOaU diffXViRQ. FRU diOXWe
VROXWiRQV Rf a ViQgOe VaOW, Whe diffXViRQ cRefficieQW iV giYeQ b\ Whe NeUQVW-HaVNeOO eTXaWiRQ (HaUQed,
H. S., aQd B. B. OZeQ, "The Ph\VicaO ChePiVWU\ Rf EOecWURO\Wic SROXWiRQV," ACS MRQRgU. 95, 1950):
a) EVWiPaWe Whe diffXViRQ cRefficieQW aW 298 K fRU a YeU\ diOXWe VROXWiRQ Rf HCO iQ ZaWeU.
SROXWiRQ
b) EVWiPaWe Whe diffXViRQ cRefficieQW aW 273 K fRU a YeU\ diOXWe VROXWiRQ Rf CXSO4 iQ ZaWeU. The
YiVcRViW\ Rf OiTXid ZaWeU aW 273 K iV 1.79 cP.
1.21a. O[\gen diffusion in Zater: 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 u 10ç5 cm2/s (Cussler E. L., DiffXsion, 2nd ed, Cambridge
University Press, Cambridge, UK, 1997). 
Solution
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 u 10ç5 cm2/s. The following data are available (Reid, et al., 1987):
Solution
1.23b. EVWimaWing molaU YolXmeV fUom liTXid diffXVion daWa.
The diffusivity of allyl alcohol (C3H6O) in dilute aqueous solution at 288 K is 0.9 u 10
ç5 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
From Table 2.1
1.24b, d. Concentration dependence of binar\ liquid diffusivities.
a) EVWiPaWe Whe diffXViYiW\ Rf eWhaQRO iQ ZaWeU aW 298 K ZheQ Whe PRO fUacWiRQ Rf eWhaQRO iQ VROXWiRQ iV
40%. UQdeU WheVe cRQdiWiRQV (HaPPRQd, B. R., aQd R. H. SWRNeV, Trans. Farada\ Soc., 49, 890,
1953):
The e[SeUiPeQWaO YaOXe UeSRUWed b\ HaPPRQd aQd SWRNeV (1953) iV 0.42 u 10-5
cP2/V.
SROXWiRQ
EVWiPaWe Whe iQfiQiWe diOXWiRQ diffXViYiW\ Rf eWhaQRO iQ ZaWeU aW 298 K
fURP Ha\dXN-MiQhaV fRU aTXeRXV VROXWiRQV
FURP ASSeQdi[ A, Whe iQfiQiWe diOXWiRQ diffXViYiW\ Rf ZaWeU iQ eWhaQRO
aW 298 K iV
b) EVWiPaWe Whe diffXViYiW\ Rf aceWRQe iQ ZaWeU aW 298 K ZheQ Whe PRO fUacWiRQ Rf aceWRQe iQ VROXWiRQ
iV 35%. FRU WhiV V\VWeP aW 298 K, Whe acWiYiW\ cRefficieQW fRU aceWRQe iV giYeQ b\ WiOVRQ eTXaWiRQ
(SPiWh, J. M., eW aO., Introduction to Chemical Engineering Thermod\namics, 5Wh ed, McGUaZ-HiOO
CR., IQc., NeZ YRUN, NY, 1996):
SROXWiRQ
EVWiPaWe Whe WheUPRd\QaPic facWRU
FURP ASSeQdi[ A
EVWiPaWe Whe iQfiQiWe diOXWiRQ diffXViYiW\ Rf aceWRQe iQ ZaWeU aW 298 K
fURP Ha\dXN-MiQhaV fRU aTXeRXV VROXWiRQV
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)
Calculate binar\ MS diffusivities from Wilke-Lee
Appendi[ C-2: Solution of the Ma[Zell-Stefan equations for a
multicomponent mi[ture of ideal gases b\ orthogonal collocation (1C =
3).
Orthogonal collocation matrices
The pUeVVXUe and WempeUaWXUe in Whe YapoXU phaVe aUe
The Ma[Zell-SWefan diffXVion coefficicienWV aUe
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)
1.26b. Stead\-state, one-dimensional, liquid-phase flu[ calculation.
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YLUWXDOO\ SXUH ZDWHU. TKH VROXELOLW\ RI GODXEHU'V VDOW LQ ZDWHU DW 288 K LV 36 J RI FU\VWDO/100 J RI
ZDWHU DQG WKH GHQVLW\ RI WKH FRUUHVSRQGLQJ VDWXUDWHG VROXWLRQ LV 1,240 NJ/P3 (PHUU\ DQG CKLOWRQ,
1973). TKH GLIIXVLYLW\ RI ND2SO4 LQ GLOXWH DTXHRXV VROXWLRQ DW 288 K FDQ EH HVWLPDWHG DV
VXJJHVWHG LQ PUREOHP 1-20. TKH GHQVLW\ RI SXUH OLTXLG ZDWHU DW 288 K LV 999.8 NJ/P3; WKH YLVFRVLW\
LV 1.153 FP.
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A = ND2SO4 B = H2O
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VROXWLRQ)BDVLV: 100 J H2O (36 J RI GLVVROYHG FU\VWDO)
(PXUH ZDWHU)
Calculate diffusivity
1.27c, d. MolecXlaU diffXVion WhUoXgh a gaV-liTXid inWeUface.
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 u 10ç5 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. 
SolXWion:
IniWial eVWimaWeV:
1.28c. Stead\-state molecular diffusion in gases.
A Pi[WXUe Rf eWhaQRO aQd ZaWeU YaSRU iV beiQg UecWified iQ aQ adiabaWic diVWiOOaWiRQ cROXPQ. The
aOcRhRO iV YaSRUi]ed aQd WUaQVfeUUed fURP Whe OiTXid WR Whe YaSRU ShaVe. WaWeU YaSRU cRQdeQVeV
(eQRXgh WR VXSO\ Whe OaWeQW heaW Rf YaSRUi]aWiRQ Qeeded b\ Whe aOcRhRO beiQg eYaSRUaWed) aQd iV
WUaQVfeUUed fURP Whe YaSRU WR Whe OiTXid ShaVe. BRWh cRPSRQeQWV diffXVe WhURXgh a gaV fiOP 0.1 PP
WhicN. The WePSeUaWXUe iV 368 K aQd Whe SUeVVXUe iV 1 aWP. The PROe fUacWiRQ Rf eWhaQRO iV 0.8 RQ
RQe Vide Rf Whe fiOP aQd 0.2 RQ Whe RWheU Vide Rf Whe fiOP. CaOcXOaWe Whe UaWe Rf diffXViRQ Rf eWhaQRO
aQd Rf ZaWeU, iQ Ng/P2-V. The OaWeQW heaW Rf YaSRUi]aWiRQ Rf Whe aOcRhRO aQd ZaWeU aW 368 K caQ be
eVWiPaWed b\ Whe PiW]eU aceQWUic facWRU cRUUeOaWiRQ (Reid, eW aO., 1987)
ZheUe Z iV Whe aceQWUic facWRU.
SROXWiRQ 
A = eWhaQRO B = ZaWeU
CaOcXOaWe eWhaQRO heaW Rf YaSRUi]aWiRQ
CaOcXOaWe ZaWeU heaW Rf YaSRUi]aWiRQ
EVWiPaWe diffXViYiW\ fURP WiONe-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 u 10
-5
kg/m-s, and r = 1.18 kg/m3.
Solution
Estimate diffusivit\ from the Wilke-Lee equation
1.30b, d. Stead\-state molecular diffusion in gases.
WaWeU eYaSRUaWiQg fURP a SRQd aW 300 K dReV VR b\ PROecXOaU diffXViRQ acURVV aQ aiU fiOP 1.5 PP
WhicN. If Whe UeOaWiYe hXPidiW\ Rf Whe aiU aW Whe RXWeU edge Rf Whe fiOP iV 20%, aQd Whe WRWaO SUeVVXUe iV 1
baU, eVWiPaWe Whe dURS iQ Whe ZaWeU OeYeO SeU da\, aVVXPiQg WhaW cRQdiWiRQV iQ Whe fiOP UePaiQ
cRQVWaQW. The YaSRU SUeVVXUe Rf ZaWeU aV a fXQcWiRQ Rf WePSeUaWXUe caQ be accXUaWeO\ eVWiPaWed fURP
Whe WagQeU eTXaWiRQ (Reid, eW aO., 1987)
SROXWiRQ
FURP ASSeQdi[ A
1.31b, d. Stead\-state molecular diffusion in a ternar\ gas s\stem.
CaOcXOaWe Whe fOX[eV aQd cRQceQWUaWiRQ SURfiOeV fRU Whe WeUQaU\ V\VWeP h\dURgeQ (1), QiWURgeQ (2), aQd
caUbRQ diR[ide (3) XQdeU Whe fROORZiQg cRQdiWiRQV. The WePSeUaWXUe iV 308 K aQd Whe SUeVVXUe iV 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, 1 = 0. The MS diffusion coefficients are D12 = 83.8 mm
2/s, D13 = 68.0
mm2/s, D23 = 16.8 mm
2/s.
Solution
Appendi[ C-1: Solution of the Ma[Zell-Stefan equations for a
multicomponent mi[ture of ideal gases b\ orthogonal collocation (1C =
3).
Orthogonal collocation matrices
The pUeVVXUe and WempeUaWXUe in Whe YapoXU phaVe aUe
The Ma[Zell-SWefan diffXVion coefficicienWV aUe
The lengWh of Whe diffXVion paWh iV
The denViW\ of Whe gaV phaVe folloZV fUom Whe ideal gaV laZ
Initial estimates of the flu[es
Initial estimates of the concentrations
2.1a. MaVV-WUanVfeU coefficienWV in a gaV abVoUbeU.
A gaV abVRUbeU iV XVed WR UePRYe beQ]eQe (C6H6) YaSRUV fURP aiU b\ VcUXbbiQg Whe gaV Pi[WXUe
ZiWh a QRQYROaWiOe RiO aW 300 K aQd 1 aWP. AW a ceUWaiQ SRiQW iQ Whe abVRUbeU, Whe beQ]eQe PROe
fUacWiRQ iQ Whe bXON Rf Whe gaV ShaVe iV 0.02, ZhiOe Whe cRUUeVSRQdiQg iQWeUfaciaO beQ]eQe gaV-
ShaVe cRQceQWUaWiRQ iV 0.0158. The beQ]eQe fOX[ aW WhaW SRiQW iV PeaVXUed aV 0.62 g/P2-V. 
a) CaOcXOaWe Whe PaVV-WUaQVfeU cRefficieQW iQ Whe gaV ShaVe aW WhaW SRiQW iQ Whe eTXiSPeQW,
e[SUeVViQg Whe dUiYiQg fRUce iQ WeUPV Rf PROe fUacWiRQV .
SROXWiRQ
b) CaOcXOaWe Whe PaVV-WUaQVfeU cRefficieQW iQ Whe gaV ShaVe aW WhaW SRiQW iQ Whe eTXiSPeQW,
e[SUeVViQg Whe dUiYiQg fRUce iQ WeUPV Rf PROaU cRQceQWUaWiRQV, NPRO/P3.
SROXWiRQ
c) AW Whe VaPe SOace iQ Whe eTXiSPeQW, Whe beQ]eQe PROe fUacWiRQ iQ Whe bXON Rf Whe OiTXid ShaVe
iV 0.125, ZhiOe Whe cRUUeVSRQdiQg iQWeUfaciaO beQ]eQe OiTXid-ShaVe cRQceQWUaWiRQ iV 0.158.
CaOcXOaWe Whe PaVV-WUaQVfeU cRefficieQW iQ Whe OiTXid ShaVe, e[SUeVViQg Whe dUiYiQg fRUce iQ WeUPV Rf
PROe fUacWiRQV.
SROXWiRQ
2.2a. MaVV-WUanVfeU coefficienWV fUom naphWhalene VXblimaWion daWa.
IQ a OabRUaWRU\ e[SeUiPeQW, aiU aW 347 K aQd 1 aWP iV bORZQ aW high VSeed aURXQd a ViQgOe
QaShWhaOeQe (C10H8) VSheUe, Zhich VXbOiPaWeV SaUWiaOO\. WheQ Whe e[SeUiPeQW begiQV, Whe diaPeWeU
Rf Whe VSheUe iV 2.0 cP. AW Whe eQd Rf Whe e[SeUiPeQW, 14.32 PiQ OaWeU, Whe diaPeWeU Rf Whe VSheUe iV
1.85 cP. 
a) EVWiPaWe Whe PaVV-WUaQVfeU cRefficieQW, baVed RQ Whe aYeUage VXUface aUea Rf Whe SaUWicOe,
e[SUeVViQg Whe dUiYiQg fRUce iQ WeUPV Rf SaUWiaO SUeVVXUeV. The deQViW\ Rf VROid QaShWhaOeQe iV 1.145
g/cP3, iWV YaSRU SUeVVXUe aW 347 K iV 670 Pa (PeUU\ aQd ChiOWRQ, 1973).
SROXWiRQ
b) CaOcXOaWe Whe PaVV-WUaQVfeU cRefficieQW, fRU Whe dUiYiQg fRUce iQ WeUPV Rf PROaU cRQceQWUaWiRQV.
SROXWiRQ
2.3a. Mass-transfer coefficients from acetone eYaporation data.
IQ a OabRUaWRU\ e[SeUiPeQW, aiU aW 300 K aQd 1 aWP iV bORZQ aW high VSeed SaUaOOeO WR Whe VXUface Rf a
UecWaQgXOaU VhaOORZ SaQ WhaW cRQWaiQV OiTXid aceWRQe(C3H6O), Zhich eYaSRUaWeV SaUWiaOO\. The SaQ iV
1 P ORQg aQd 50 cPV Zide. IW iV cRQQecWed WR a UeVeUYRiU cRQWaiQiQg OiTXid aceWRQe Zhich
aXWRPaWicaOO\ UeSOaceV Whe aceWRQe eYaSRUaWed, PaiQWaiQiQg a cRQVWaQW OiTXid OeYeO iQ Whe SaQ. DXUiQg
aQ e[SeUiPeQWaO UXQ, iW ZaV RbVeUYed WhaW 2.0 L Rf aceWRQe eYaSRUaWed iQ 5 PiQ. EVWiPaWe Whe PaVV-
WUaQVfeU cRefficieQW. The deQViW\ Rf OiTXid aceWRQe aW 300 K iV 0.79 g/cP3; iWV YaSRU SUeVVXUe iV 27
NPa (PeUU\ aQd ChiOWRQ, 1973).
SROXWiRQ
2.4b. Mass-transfer coefficients from Zetted-Zall e[perimental data.
A ZeWWed-ZaOO e[SeUiPeQWaO VeW-XS cRQViVWV Rf a gOaVV SiSe, 50 PP iQ diaPeWeU aQd 1.0 P ORQg.
WaWeU aW 308 K fORZV dRZQ Whe iQQeU ZaOO. DU\ aiU eQWeUV Whe bRWWRP Rf Whe SiSe aW Whe UaWe Rf 1.04
P3/PiQ, PeaVXUed aW 308 K aQd 1 aWP. IW OeaYeV Whe ZeWWed VecWiRQ aW 308 K aQd ZiWh a UeOaWiYe
hXPidiW\ Rf 34%. WiWh Whe heOS Rf eTXaWiRQ (2-52), eVWiPaWe Whe aYeUage PaVV-WUaQVfeU cRefficieQW,
ZiWh Whe dUiYiQg fRUce iQ WeUPV Rf PROaU fUacWiRQV.
SROXWiRQ 
2.7c. MaVV WUanVfeU in an annXlaU Vpace.
a) IQ VWXd\iQg UaWeV Rf diffXViRQ Rf QaShWhaOeQe iQWR aiU, aQ iQYeVWigaWRU UeSOaced a 30.5-cP VecWiRQ Rf
Whe iQQeU SiSe Rf aQ aQQXOXV ZiWh a QaShWhaOeQe URd. The aQQXOXV ZaV cRPSRVed Rf a 51-PP-OD
bUaVV iQQeU SiSe VXUURXQded b\ a 76-PP-ID bUaVV SiSe. WhiOe RSeUaWiQg aW a PaVV YeORciW\ ZiWhiQ Whe
aQQXOXV Rf 12.2 Ng Rf aiU/P2-V aW 273 K aQd 1 aWP, Whe iQYeVWigaWRU deWeUPiQed WhaW Whe SaUWiaO
SUeVVXUe Rf QaShWhaOeQe iQ Whe e[iWiQg gaV VWUeaP ZaV 0.041 Pa. UQdeU Whe cRQdiWiRQV Rf Whe
iQYeVWigaWiRQ, Whe SchPidW QXPbeU Rf Whe gaV ZaV 2.57, Whe YiVcRViW\ ZaV 175 PP, aQd Whe YaSRU
SUeVVXUe Rf QaShWhaOeQe ZaV 1.03 Pa. EVWiPaWe Whe PaVV-WUaQVfeU cRefficieQW fURP Whe iQQeU ZaOO fRU
WhiV VeW Rf cRQdiWiRQV. AVVXPe WhaW eTXaWiRQ (2-52) aSSOieV.
SROXWiRQ
b) MRQUad aQd PeOWRQ (TUanV. AIChE, 38, 593, 1942) SUeVeQWed Whe fROORZiQg cRUUeOaWiRQ fRU
heaW-WUaQVfeU cRefficieQW iQ aQ aQQXOaU VSace:
ZheUe dR aQd di aUe Whe RXWVide aQd iQVide diaPeWeUV Rf Whe aQQXOXV, de iV Whe eTXiYaOeQW
diaPeWeU defiQed aV 
WUiWe dRZQ Whe aQaORgRXV e[SUeVViRQ fRU PaVV WUaQVfeU aQd XVe iW WR eVWiPaWe Whe PaVV-WUaQVfeU
cRefficieQW fRU Whe cRQdiWiRQV Rf SaUW a). CRPSaUe bRWh UeVXOWV.
SROXWiRQ
2.8c. The Chilton-Colburn analog\: flow across tube banks.
WLQGLQJ DQG CKHQH\ (IQd. EQg. CheP., 40, 1087, 1948) SDVVHG DLU DW 310 K DQG 1 DWP WKURXJK D
EDQN RI URGV RI QDSKWKDOHQH. TKH URGV ZHUH LQ D VWDJJHUHG DUUDQJHPHQW, ZLWK WKH DLU IORZLQJ DW ULJKW
DQJOHV WR WKH D[HV RI WKH URGV. TKH EDQN FRQVLVWHG RI 10 URZV FRQWDLQLQJ DOWHUQDWHO\ ILYH DQG IRXU
38-PP-OD WXEHV (G = 38 PP) VSDFHG RQ 57-PP FHQWHUV, ZLWK WKH URZV 76 PP DSDUW. TKH PDVV-
WUDQVIHU FRHIILFLHQW ZDV GHWHUPLQHG E\ PHDVXULQJ WKH UDWH RI VXEOLPDWLRQ RI WKH QDSKWKDOHQH. TKH
GDWD FRXOG EH FRUUHODWHG E\:
ZKHUH G' LV WKH PD[LPXP PDVV YHORFLW\ WKURXJK WKH WXEH EDQN, LQ NJ/P2-V, DQG NG LV LQ NPRO/P
2-
V-PD.
D) RHZULWH HTXDWLRQ (2-68) LQ WHUPV RI WKH CROEXUQ MD -IDFWRU. TKH GLIIXVLYLW\ RI QDSKWKDOHQH LQ DLU DW
310 K DQG 1 DWP LV 0.074 FP2/V.
SROXWLRQ
PURSHUWLHV RI GLOXWH PL[WXUHV RI QDSKWKDOHQH LQ DLU DW 310 K DQG 1 DWP:
DLPHQVLRQDO FRQVWDQW LQ WKH JLYHQ FRUUHODWLRQ:
b) EVWiPaWe Whe PaVV-WUaQVfeU cRefficieQW WR be e[SecWed fRU eYaSRUaWiRQ Rf Q-SURS\O aOcRhRO iQWR
caUbRQ diR[ide fRU Whe VaPe geRPeWUicaO aUUaQgePeQW ZheQ Whe caUbRQ diR[ide fORZV aW a Pa[iPXP
YeORciW\ Rf 10 P/V aW 300 K aQd 1 aWP. The YaSRU SUeVVXUe Rf Q-SURS\O aOcRhRO aW 300 K iV 2.7 NPa.
SROXWiRQ
PURSeUWieV Rf diOXWe Pi[WXUeV Rf SURS\O aOcRhRO iQ caUbRQ diR[ide aW 300 K aQd 1 aWP:
c) ZaNaXVNaV (AdY. Heat Transfer, �, 93, 1972) SURSRVed Whe fROORZiQg cRUUeOaWiRQ fRU Whe heaW-
WUaQVfeU cRefficieQW iQ a VWaggeUed WXbe baQN aUUaQgePeQW ViPiOaU WR WhaW VWXdied b\ WiQdiQg aQd
CheQe\:
UVe Whe PaVV-WUaQVfeU e[SUeVViRQ aQaORgRXV WR eTXaWiRQ (2-69) WR eVWiPaWe Whe PaVV-WUaQVfeU
cRefficieQW Rf SaUW b). CRPSaUe Whe UeVXOWV.
SROXWiRQ
2.9b. MaVV WUanVfeU fUom a flaW plaWe.
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. MaVV WUanVfeU fUom a flaW plaWe.
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 u 10ç6 m2/s, and the diffusivity is
1.25 u 10ç9 m2/s. The solubility of NaCl in water at 291 K is 0.35 g/cm3, and the density of the
saturated solution is 1.22 g/cm3 (Perry and Chilton, 1973) .
Solution
Laminar floZ
At the bulk of the solution, point 2:
At the interface, point 1:
2.11b. MaVV WUanVfeU fUom a flaW liTXid VXUface.
DXUiQg Whe e[SeUiPeQW deVcUibed iQ PURbOeP 2.3, Whe aiU YeORciW\ ZaV PeaVXUed aW 6 P/V, SaUaOOeO WR
Whe ORQgeVW Vide Rf Whe SaQ. EVWiPaWe Whe PaVV-WUaQVfeU cRefficieQW SUedicWed b\ eTXaWiRQ (2-28) RU
(2-29) aQd cRPSaUe iW WR Whe YaOXe PeaVXUed e[SeUiPeQWaOO\. NRWice WhaW, dXe WR Whe high YROaWiOiW\ Rf
aceWRQe, Whe aYeUage aceWRQe cRQceQWUaWiRQ iQ Whe gaV fiOP iV UeOaWiYeO\ high. TheUefRUe, SURSeUWieV
VXch aV deQViW\ aQd YiVcRViW\ VhRXOd be eVWiPaWed caUefXOO\. The fROORZiQg daWa fRU aceWRQe PighW
be Qeeded: Tc = 508.1 K, Pc = 47.0 baU, M = 58, Vc = 209 cP
3/PRO, Zc = 0.232 (Reid, eW aO., 1987).
SROXWiRQ
AYeUage fiOP SURSeUWieV:
EVWiPaWe Whe YiVcRViW\ Rf Whe Pi[WXUe fURP LXcaV MeWhRd
EVWiPaWe Whe diffXViYiW\ fURP Whe WiONe-Lee eTXaWiRQ
2.12b. EYaporation of a drop of Zater falling in air.
Repeat E[ample 2.9 for a drop of Zater Zhich is originall\ 2 mm in diameter.
Solution
2.13b. Dissolution of a solid sphere into a floZing 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 u 10ç6 m2/s, and
the mass diffusivity is 1.25 u 10ç9 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
Lennard-Jones parameters for naphthalene
2.15b. Dissolution of a solid sphere into a floZing 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
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 Lleads 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 u 10ç5 cm2/s
(Cussler, 1997).
Solution
2.17b. Mass transfer in a Zetted-Zall toZer.
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
P ORQg.
SROXWiRQ
FRU ZaWeU aW 294 K
2.18c. MaVV WUanVfeU in an annXlaU Vpace.
IQ VWXd\iQg Whe VXbOiPaWiRQ Rf QaShWhaOeQe iQWR aQ aiUVWUeaP, aQ iQYeVWigaWRU cRQVWUXcWed a 3-P-ORQg
aQQXOaU dXcW. The iQQeU SiSe ZaV Pade fURP a 25-PP-OD, VROid QaShWhaOeQe URd; WhiV ZaV
VXUURXQded b\ a 50-PP-ID QaShWhaOeQe SiSe. 
AiU aW 289 K aQd 1 aWP fORZed WhURXgh Whe aQQXOaU VSace aW aQ aYeUage YeORciW\ Rf 15 P/V. EVWiPaWe
Whe SaUWiaO SUeVVXUe Rf QaShWhaOeQe iQ Whe aiUVWUeaP e[iWiQg fURP Whe WXbe. AW 289 K, QaShWhaOeQe
haV a YaSRU SUeVVXUe Rf 5.2 Pa, aQd a diffXViYiW\ iQ aiU Rf 0.06 cP2/V. UVe Whe UeVXOWV Rf PURbOeP 2.7
WR eVWiPaWe Whe PaVV-WUaQVfeU cRefficieQW fRU Whe iQQeU VXUface; aQd eTXaWiRQ (2-47), XViQg Whe
eTXiYaOeQW diaPeWeU defiQed iQ PURbOeP 2.7, WR eVWiPaWe Whe cRefficieQW fURP Whe RXWeU VXUface.
SolXWion
In WhiV ViWXaWion, WheUe Zill be a molaU flX[ fUom Whe inneU Zall, NA1, ZiWh Vpecific inWeUfacial aUea, a1,
and a flX[ fUom Whe oXWeU Zall, NA2, ZiWh aUea a2. A maWeUial balance on a diffeUenWial YolXme
elemenW \ieldV:
Define:
Then:
FoU Whe inWeUioU Zall:
For the outer Zall:
2.19c. Ben]ene evaporation on the outside surface of a single c\linder.
BeQ]eQe iV eYaSRUaWiQg aW Whe UaWe Rf 20 Ng/hU RYeU Whe VXUface Rf a SRURXV 10-cP-diaPeWeU c\OiQdeU.
DU\ aiU aW 325 K aQd 1 aWP fORZV aW UighW aQgOe WR Whe a[iV Rf Whe c\OiQdeU aW a YeORciW\ Rf 2 P/V. The
OiTXid iV aW a WePSeUaWXUe Rf 315 K ZheUe iW e[eUWV a YaSRU SUeVVXUe Rf 26.7 NPa. EVWiPaWe Whe OeQgWh
Rf Whe c\OiQdeU. FRU beQ]eQe, Tc = 562.2 K, Pc = 48.9 baU, M = 78, Vc = 259 cP
3/PRO, Zc = 0.271
(Reid, eW aO., 1987).
SROXWiRQ
CaOcXOaWe Whe aYeUage SURSeUWieV Rf Whe fiOP
FURP Whe WiONe-Lee eTXaWiRQ
FURP Whe LXcaV MeWhRd
FURP ET. 2-45:
2.20b. MaVV WUanVfeU in a packed bed.
Wilke and Hougan (TUanV. 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 u 10
ç3 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
From the Wilke-Lee equation
2.21b. MaVV WUanVfeU and pUeVVXUe dUop 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
From the Wilke-Lee equation:
b) EVWimaWe Whe pUeVVXUe dUop WhUoXgh Whe bed.
SolXWion
2.22b. Volumetric mass-transfer coefficients in industrial toZers.
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 u 10
ç9 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/m
2-s. For the conditions described in part a), estimate the liquid-film
mass-transfer coefficient using equation (2-72). Compare the results.
Solution
2.23b. Mass transfer in fluidi]ed beds.
Cavatorta, et al. (AICKE J., 45, 938, 1999) studied the electrochemical reduction of ferrycianide
ions, {Fe(CN)6}
ç3, to ferrocyanide, {Fe(CN)6}
ç4, 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 u 10ç10 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%.
Solution
b) CaYaWoUWa eW al. pUopoVed Whe folloZing coUUelaWion Wo eVWimaWe Whe maVV-WUanVfeU coefficienW foU
WheiU flXidi]ed bed e[peUimenWal UXnV:
ZheUe Re iV baVed on Whe empW\ WXbe YelociW\. UVing WhiV coUUelaWion, eVWimaWe Whe maVV-WUanVfeU
coefficienW, NL, if Whe poUoViW\ of Whe bed iV 60%. CompaUe \oXU UeVXlW Wo WhaW of paUW a).
SolXWion
2.24b. Mass transfer in a holloZ-fiber boiler feedZater deaerator.
ConVideU Whe holloZ-fibeU BFW deaeUaWoU deVcUibed in E[ample 2-13. If Whe ZaWeU floZ UaWe
incUeaVeV Wo 60,000 kg/hU Zhile eYeU\Whing elVe UemainV conVWanW, calcXlaWe Whe fUacWion of Whe
enWeUing diVVolYed o[\gen WhaW can be UemoYed.
SolXWion
2.25b. Mass transfer in a holloZ-fiber boiler feedZater deaerator.
a) ConVideU Whe holloZ-fibeU BFW deaeUaWoU deVcUibed in E[ample 2-13. AVVXming WhaW onl\ o[\gen
diffXVeV acUoVV Whe membUane, calcXlaWe Whe gaV YolXme floZ UaWe and compoViWion aW Whe lXmen
oXWleW. The ZaWeU enWeUV Whe Vhell Vide aW 298 K VaWXUaWed ZiWh aWmoVpheUic o[\gen, Zhich meanV a
diVVolYed o[\gen concenWUaWion of 8.38 mg/L.
SolXWion
b) CaOcXOaWe Whe PaVV-WUaQVfeU cRefficieQW aW Whe aYeUage cRQdiWiRQV iQVide Whe OXPeQ. NegOecW Whe
WhicNQeVV Rf Whe fibeU ZaOOV ZheQ eVWiPaWiQg Whe gaV YeORciW\ iQVide Whe OXPeQ.
SROXWiRQ
CaOcXOaWe Whe aYeUage fORZ cRQdiWiRQV iQVide Whe fibeUV
CaOcXOaWe Whe aYeUage R[\geQ PROaU fUacWiRQ iQ Whe gaV
FURP LXcaV PeWhRd fRU SXUe N2
FURP Whe WiONe-Lee eTXaWiRQ
(From E[ample 2.13)
3.1a. Application of Raoult's law to a binar\ s\stem.
ReSeaWE[aPSOe 3.1, bXW fRU a OiTXid cRQceQWUaWiRQ Rf 0.6 PROe fUacWiRQ Rf beQ]eQe aQd a
WePSeUaWXUe Rf 320 K.
SROXWiRQ
3.2b. Application of Raoult's law to a binar\ s\stem.
a) DeWeUPiQe Whe cRPSRViWiRQ Rf Whe OiTXid iQ eTXiOibUiXP ZiWh a YaSRU cRQWaiQiQg 60 PROe SeUceQW
beQ]eQe-40 PROe SeUceQW WROXeQe if Whe V\VWeP e[iVWV iQ a YeVVeO XQdeU 1 aWP SUeVVXUe. PUedicW
Whe eTXiOibUiXP WePSeUaWXUe.
SROXWiRQ
IQiWiaO eVWiPaWeV
b) DeWeUPiQe Whe cRPSRViWiRQ Rf Whe YaSRU iQ eTXiOibUiXP ZiWh a OiTXid cRQWaiQiQg 60 PROe SeUceQW
beQ]eQe-40 PROe SeUceQW WROXeQe if Whe V\VWeP e[iVWV iQ a YeVVeO XQdeU 1 aWP SUeVVXUe. PUedicW
Whe eTXiOibUiXP WePSeUaWXUe.
SROXWiRQ
3.3a. Application of Raoult's laZ to a binar\ s\stem.
NRUPaO heSWaQe, Q-C7H16, aQd QRUPaO RcWaQe, Q-C8H18, fRUP ideaO VROXWiRQV. AW 373 K, QRUPaO
heSWaQe haV a YaSRU SUeVVXUe Rf 106 NPa aQd QRUPaO RcWaQe Rf 47.1 NPa.
a) WhaW ZRXOd be Whe cRPSRViWiRQ Rf a heSWaQe-RcWaQe VROXWiRQ WhaW bRiOV aW 373 K XQdeU a 93 NPa
SUeVVXUe?
SROXWiRQ
b) WhaW ZRXOd be Whe cRPSRViWiRQ Rf Whe YaSRU iQ eTXiOibUiXP ZiWh Whe VROXWiRQ WhaW iV deVcUibed iQ
(a)?
SROXWiRQ
3.4a. Henr\'s laZ: saturation of Zater Zith o[\gen.
A VROXWiRQ ZiWh R[\geQ diVVROYed iQ ZaWeU cRQWaiQiQg 0.5 Pg O2/100 g Rf H2O iV bURXghW iQ cRQWacW
ZiWh a OaUge YROXPe Rf aWPRVSheUic aiU aW 283 K aQd a WRWaO SUeVVXUe Rf 1 aWP. The HeQU\'V OaZ
cRQVWaQW fRU Whe R[\geQ-ZaWeU V\VWeP aW 283 K eTXaOV 3.27 u 104 aWP/PROe fUacWiRQ.
a) WiOO Whe VROXWiRQ gaiQ RU ORVe R[\geQ?
b) WhaW ZiOO be Whe cRQceQWUaWiRQ Rf R[\geQ iQ Whe fiQaO eTXiOibUiXP VROXWiRQ?
SROXWiRQ
AW eTXiOibUiXP:
BaViV: 1 L ZaWeU (1 Ng ZaWeU)
ETXiOibUiXP cRQceQWUaWiRQ, ce = 11.42 Pg R[\geQ/L
IQiWiaO cRQdiWiRQV:
The VROXWiRQ gaiQV R[\geQ.
3.5c. Material balances combined Zith equilibrium relations.
ReSeaW E[aPSOe 3.3, bXW aVVXPiQg WhaW Whe aPPRQia, aiU, aQd ZaWeU aUe bURXghW iQWR cRQWacW iQ a
cORVed cRQWaiQeU. TheUe iV 10 P3 Rf gaV VSace RYeU Whe OiTXid. AVVXPiQg WhaW Whe gaV-VSace YROXPe
aQd Whe WePSeUaWXUe UePaiQ cRQVWaQW XQWiO eTXiOibUiXP iV achieYed, PRdif\ Whe MaWhcad SURgUaP iQ
FigXUe 3.2 WR caOcXOaWe:
a) Whe WRWaO SUeVVXUe aW eTXiOibUiXP
SROXWiRQ
IQiWiaO gXeVVeV
AQVZHU: P = 1.755 DWP
E) WKH HTXLOLEULXP DPPRQLD FRQFHQWUDWLRQ LQ WKH JDV DQG OLTXLG SKDVHV.
AQVZHU: \A = 0.145; [A=
0.162
3.6b. MaVV-WUanVfeU UeViVWanceV dXUing abVoUpWion.
IQ WKH DEVRUSWLRQ RI FRPSRQHQW A (PROHFXODU ZHLJKW = 60) IURP DQ DLUVWUHDP LQWR DQ DTXHRXV
VROXWLRQ, WKH EXON FRPSRVLWLRQV RI WKH WZR DGMDFHQW VWUHDPV DW D SRLQW LQ WKH DSSDUDWXV ZHUH
DQDO\]HG WR EH SA,G = 0.1 DWP, DQG FA,L = 1.0 NPRO RI A/P
3 RI VROXWLRQ. TKH WRWDO SUHVVXUH ZDV 2.0
DWP; WKH GHQVLW\ RI WKH VROXWLRQ ZDV 1,100 NJ/P3. TKH HHQU\'V FRQVWDQW IRU WKHVH FRQGLWLRQV ZDV
0.85 DWP/PROH IUDFWLRQ. TKH RYHUDOO JDV FRHIILFLHQW ZDV KG = 0.27 NPRO/P
2-KU-DWP. II 57% RI WKH WRWDO
UHVLVWDQFH WR PDVV WUDQVIHU UHVLGHV LQ WKH JDV ILOP, GHWHUPLQH
D) WKH JDV-ILOP FRHIILFLHQW, NG;
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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
Initial estimates of interfacial concentrations:
d) the mass flu[ of A.
Solution
Check this result b\ calculating the gas-phase flu[:
3.7b. Mass-transfer resistances during absorption.
FRU a V\VWeP iQ Zhich cRPSRQeQW A iV WUaQVfeUUiQg fURP Whe gaV ShaVe WR Whe OiTXid ShaVe, Whe
eTXiOibUiXP UeOaWiRQ iV giYeQ b\
ZheUe SA,i iV Whe eTXiOibUiXP SaUWiaO SUeVVXUe iQ aWP aQd [A,i iV Whe eTXiOibUiXP OiTXid cRQceQWUaWiRQ
iQ PROaU fUacWiRQ. AW RQe SRiQW iQ Whe aSSaUaWXV, Whe OiTXid VWUeaP cRQWaiQV 4.5 PROe % aQd Whe gaV
VWUeaP cRQWaiQV 9.0 PROe % A. The WRWaO SUeVVXUe iV 1 aWP. The iQdiYidXaO gaV-fiOP cRefficieQW aW
WhiV SRiQW iV NG = 3.0 PROe/P
2-V-aWP. FifW\ SeU ceQW Rf Whe RYeUaOO UeViVWaQce WR PaVV WUaQVfeU iV
NQRZQ WR be eQcRXQWeUed iQ Whe OiTXid ShaVe. EYaOXaWe
a) Whe RYeUaOO PaVV-WUaQVfeU cRefficieQW, K\;
SROXWiRQ
b) Whe PROaU fOX[ Rf A;
SROXWiRQ
(\Ae =
\A*)
c) The OiTXid iQWeUfaciaO cRQceQWUaWiRQ Rf A.
SROXWiRQ
3.8d. Absorption of ammonia b\ water: use of F-t\pe mass-transfer coefficients.
MRdif\ Whe MaWhcad SURgUaP iQ FigXUe 3.6 WR UeSeaW E[aPSOe 3.5, bXW ZiWh \A,G = 0.70 aQd [A,L =
0.10. EYeU\WhiQg eOVe UePaiQV cRQVWaQW.
SolXWion
IniWial gXeVVeV
3.9d. Absorption of ammonia b\ water: use of F-t\pe mass-transfer coefficients.
Modif\ Whe MaWhcad pUogUam in FigXUe 3.6 Wo UepeaW E[ample 3.5, bXW ZiWh FL = 0.0050 kmol/m
2-V.
EYeU\Whing elVe UemainV conVWanW.
SolXWion
IniWial gXeVVeV
3.10b. MaVV-WUanVfeU UeViVWanceV dXUing abVoUpWion of ammonia.
IQ Whe abVRUSWiRQ Rf aPPRQia iQWR ZaWeU fURP aQ aiU-aPPRQia Pi[WXUe a 300 K aQd 1 aWP, Whe
iQdiYidXaO fiOP cRefficieQWV ZeUe eVWiPaWed WR be NL = 6.3 cP/hU aQd NG = 1.17 NPRO/P
2-hU-aWP. The
eTXiOibUiXP UeOaWiRQVhiS fRU YeU\ diOXWe VROXWiRQV Rf aPPRQia iQ ZaWeU aW 300 K aQd 1 aWP iV
DeWeUPiQe Whe fROORZiQg PaVV-WUaQVfeU cRefficieQWV:
a) N\
SROXWiRQ
b)
N[
SROXWiRQ
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 holloZ-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, AICKE 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, AICKE 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 u 10
ç6 m (Prasad
and Sirkar, 1988).
a) Calculate the corresponding value of kM.
Solution
For oxygen in nitrogen at 298 K and 1 atm:
b) UViQg Whe UeVXOWV Rf SaUW (a), E[aPSOe 2.13, aQd PURbOeP 2.25, caOcXOaWe KL, aQd eVWiPaWe ZhaW
fUacWiRQ Rf Whe WRWaO UeViVWaQce WR PaVV WUaQVfeU UeVideV iQ Whe OiTXid fiOP.
SROXWiRQ
FURP E[aPSOe 2.13:
FURP PURb. 2.25:
ViUWXaOO\ aOO Rf Whe UeViVWaQce UeVideV iQ Whe OiTXid ShaVe.
3.12c. Combined use of F- and k-t\pe coefficients: absorption of low-solubilit\ gases.
DXUiQg abVRUSWiRQ Rf ORZ-VROXbiOiW\ gaVeV, PaVV WUaQVfeU fURP a highO\ cRQceQWUaWed gaV Pi[WXUe WR
a YeU\ diOXWe OiTXid VROXWiRQ fUeTXeQWO\ WaNeV SOace. IQ WhaW caVe, aOWhRXgh iW iV aSSURSUiaWe WR XVe a
N-W\Se PaVV-WUaQVfeU cRefficieQW iQ Whe OiTXid ShaVe, aQ F-W\Se cRefficieQW PXVW be XVed iQ Whe gaV
ShaVe. SiQce diOXWe OiTXid VROXWiRQV XVXaOO\ Rbe\ HeQU\'V OaZ, Whe iQWeUfaciaO cRQceQWUaWiRQV dXUiQg
abVRUSWiRQ Rf ORZ-VROXbiOiW\ gaVeV aUe UeOaWed WhURXgh \A,i = P[A,i.
a) ShRZ WhaW, XQdeU Whe cRQdiWiRQV deVcUibed abRYe, Whe gaV iQWeUfaciaO cRQceQWUaWiRQ VaWiVfieV Whe
eTXaWiRQ
SROXWiRQ
IQ Whe gaV ShaVe:
IQ Whe OiTXid ShaVe:
FRU HeQU\'V LaZ:
TheQ:
ReaUUaQgiQg:
b) IQ a ceUWaiQ aSSaUaWXV XVed fRU Whe abVRUSWiRQ Rf SO2 fURP aiU b\ PeaQV Rf ZaWeU, aW RQe SRiQW iQ
Whe eTXiSPeQW Whe gaV cRQWaiQed 30% SO2 b\ YROXPe aQd ZaV iQ cRQWacW ZiWh a OiTXid cRQWaiQiQg
0.2% SO2 b\ PROe. The WePSeUaWXUe ZaV 303 K aQd Whe WRWaO SUeVVXUe 1 aWP. EVWiPaWe Whe
iQWeUfaciaO cRQceQWUaWiRQV aQd Whe ORcaO SO2 PROaU fOX[. The PaVV-WUaQVfeU cRefficieQWV ZeUe
caOcXOaWed aV FG = 0.002 NPRO/P
2-V, N[ = 0.160 NPRO/P
2-V. The eTXiOibUiXP SO2 VROXbiOiW\ daWa aW
303 K aUe (PeUU\ aQd ChiOWRQ, 1973):
Ng SO2/100 Ng ZaWeU PaUWiaO SUeVVXUe Rf SO2, PP Hg (WRUU)
0.0 0
0.5 42
1.0 85
1.5 129
2.0176
2.5 224
SROXWiRQ
DefiQe: Z = Ng SO2/100 Ng ZaWeU
S = PaUWiaO SUeVVXUe Rf SO2, PP Hg (WRUU)
IniWial gXeVV:
3.13d. Distillation of a mi[ture of methanol and Zater in a packed toZer: 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 u 10
ç3 kmol/m2-s, and FL = 8.650 u 10
ç3 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)
For water (B)
PaUameWeUV
IniWial eVWimaWeV
3.14b. Material balances: adsorption of ben]ene vapor on activated carbon.
AcWiYaWed caUbRQ iV XVed WR UecRYeU beQ]eQe fURP a QiWURgeQ-beQ]eQe YaSRU Pi[WXUe. A QiWURgeQ-
beQ]eQe Pi[WXUe aW 306 K aQd 1 aWP cRQWaiQiQg 1% beQ]eQe b\ YROXPe iV WR be SaVVed
cRXQWeUcXUUeQWO\ aW Whe UaWe Rf 1.0 P3/V WR a PRYiQg VWUeaP Rf acWiYaWed caUbRQ VR aV WR UePRYe
85% Rf Whe beQ]eQe fURP Whe gaV iQ a cRQWiQXRXV SURceVV. The eQWeUiQg acWiYaWed caUbRQ cRQWaiQV
15 cP3 beQ]eQe YaSRU (aW STP) adVRUbed SeU gUaP Rf Whe caUbRQ. The WePSeUaWXUe aQd WRWaO
SUeVVXUe aUe PaiQWaiQed aW 306 K aQd 1 aWP. NiWURgeQ iV QRW adVRUbed. The eTXiOibUiXP adVRUSWiRQ
Rf beQ]eQe RQ WhiV acWiYaWed caUbRQ aW 306 K iV UeSRUWed aV fROORZV:
BeQ]eQe YaSRU adVRUbed PaUWiaO SUeVVXUe beQ]eQe, PP Hg
cP3 (STP)/g caUbRQ
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) PORW Whe eTXiOibUiXP daWa aV X' = Ng beQ]eQe/Ng dU\ caUbRQ, Y' = Ng beQ]eQe/Ng QiWURgeQ fRU a
WRWaO SUeVVXUe Rf 1 aWP.
SROXWiRQ
b) CaOcXOaWe Whe PiQiPXP fORZ UaWe UeTXiUed Rf Whe eQWeUiQg acWiYaWed caUbRQ (UePePbeU WhaW Whe
eQWeUiQg caUbRQ cRQWaiQV VRPe adVRUbed beQ]eQe).
SROXWiRQ
OQ Whe XY diagUaP, ORcaWe Whe SRiQW (X2,Y2). SiQce Whe RSeUaWiQg OiQe iV abRYe Whe eTXiOibUiXP
cXUYe aQd Whe eTXiOibUiXP cXUYe iV cRQcaYe XSZaUdV, Whe PiQiPXP RSeUaWiQg OiQe iV RbWaiQed b\
ORcaWiQg, aW Whe iQWeUVecWiRQ Rf Y = Y1 ZiWh Whe eTXiOibUiXP cXUYe, X1Pa[.
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) FRU Whe cRQdiWiRQV Rf SaUW (c), caOcXOaWe Whe QXPbeU Rf ideaO VWageV UeTXiUed.
SROXWiRQ
See VWeSZiVe cRQVWUXcWiRQ RQ Whe XY gUaSh
3.15b. Material balances: desorption of ben]ene vapor from activated carbon.
The acWiYaWed caUbRQ OeaYiQg Whe adVRUbeU Rf PURbOeP 3.14 iV UegeQeUaWed b\ cRXQWeUcXUUeQW cRQWacW
ZiWh VWeaP aW 380 K aQd 1 aWP. The UegeQeUaWed caUbRQ iV UeWXUQed WR Whe adVRUbeU, ZhiOe Whe
Pi[WXUe Rf VWeaP aQd deVRUbed beQ]eQe YaSRUV iV cRQdeQVed. The cRQdeQVaWe VeSaUaWeV iQWR aQ
RUgaQic aQd aQ aTXeRXV ShaVe aQd Whe WZR ShaVeV aUe VeSaUaWed b\ decaQWaWiRQ. DXe WR Whe ORZ
VROXbiOiW\ Rf beQ]eQe iQ ZaWeU, PRVW Rf Whe beQ]eQe ZiOO be cRQceQWUaWed iQ Whe RUgaQic ShaVe, ZhiOe
Whe aTXeRXV ShaVe ZiOO cRQWaiQ RQO\ WUaceV Rf beQ]eQe. The eTXiOibUiXP adVRUSWiRQ daWa aW 380 K
aUe aV fROORZV:
BeQ]eQe YaSRU adVRUbed PaUWiaO SUeVVXUe beQ]eQe, NPa
Ng beQ]eQe/100 Ng caUbRQ
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) CaOcXOaWe Whe PiQiPXP VWeaP fORZ UaWe UeTXiUed.
SROXWiRQ ETXiOibUiXP cXUYe
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:
3.17b. Material balances in batch processes: dr\ing of soap with air.
IW iV deViUed WR dU\ 10 Ng Rf VRaS fURP 20% PRiVWXUe b\ ZeighW WR QR PRUe WhaQ 6% PRiVWXUe b\
cRQWacW ZiWh hRW aiU. The ZeW VRaS iV SOaced iQ a YeVVeO cRQWaiQiQg 8.06 P3 Rf aiU aW 350 K, 1 aWP,
aQd a ZaWeU-YaSRU SaUWiaO SUeVVXUe Rf 1.6 NPa. The V\VWeP iV aOORZed WR Ueach eTXiOibUiXP, aQd
WheQ Whe aiU iQ Whe YeVVeO iV eQWiUeO\ UeSOaced b\ fUeVh aiU Rf Whe RUigiQaO PRiVWXUe cRQWeQW aQd
WePSeUaWXUe. HRZ PaQ\ WiPeV PXVW Whe SURceVV be UeSeaWed iQ RUdeU WR Ueach Whe VSecified VRaS
PRiVWXUe cRQWeQW Rf QR PRUe WhaQ 6%? WheQ WhiV VRaS iV e[SRVed WR aiU aW 350 K aQd 1 aWP, Whe
eTXiOibUiXP diVWUibXWiRQ Rf PRiVWXUe beWZeeQ Whe aiU aQd Whe VRaS iV aV fROORZV:
WW % PRiVWXUe iQ VRaS PaUWiaO SUeVVXUe Rf ZaWeU, NPa
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
SROXWiRQ
GeQeUaWe Whe XY diagUaP
From the XY diagram, at the e[it of the fifth equilibrium stage, X = 0.06 and
3.18b. Material balances in batch processes: e[traction of an aqueous nicotine solution
Zith kerosene.
Nicotine in a Zater solution containing 2% nicotine is to be e[tracted Zith 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 folloZs (Claffe\ et al., IQd. EQg. CheP.,
42, 166, 1950):
X', u 103 kg nicotine/kg Zater Y', u 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
From the XY diagram, after 4 e[tractions, X = 0.00422
3.19b. Cross-floZ cascade of ideal stages.
The dU\iQg aQd OiTXid-OiTXid e[WUacWiRQ RSeUaWiRQV deVcUibed iQ PURbOePV 3.17 aQd 3.18,
UeVSecWiYeO\, aUe e[aPSOeV Rf a fORZ cRQfigXUaWiRQ caOOed a cURVV-fORZ caVcade. FigXUe 3.27 iV a
VchePaWic diagUaP Rf a cURVV-fORZ caVcade Rf ideaO VWageV. Each VWage iV UeSUeVeQWed b\ a ciUcOe,
aQd ZiWhiQ each VWage PaVV WUaQVfeU RccXUV aV if iQ cRcXUUeQW fORZ. The L ShaVe fORZV fURP RQe
VWage WR Whe Qe[W, beiQg cRQWacWed iQ each VWage b\ a fUeVh V ShaVe. If Whe eTXiOibUiXP-diVWUibXWiRQ
cXUYe Rf Whe cURVV-fORZ caVcade iV eYeU\ZheUe VWUaighW aQd Rf VORSe P, iW caQ be VhRZQ WhaW
(TUe\baO, 1980)
ZheUe S iV Whe VWUiSSiQg facWRU, PVS/LS, cRQVWaQW fRU aOO VWageV, aQd N iV Whe WRWaO QXPbeU Rf
VWageV.
SROYe PURbOeP 3.18 XViQg eTXaWiRQ (3-60), aQd cRPSaUe Whe UeVXOWV RbWaiQed b\ Whe WZR PeWhRdV.
SROXWiRQ
IQiWiaO eVWiPaWe
3.20a. Cross-floZ cascade of ideal stages: nicotine e[traction.
CRQVideU Whe QicRWiQe e[WUacWiRQ Rf PURbOePV 3.18 aQd 3.19. CaOcXOaWe Whe QXPbeU Rf ideaO VWageV
UeTXiUed WR achieYe aW OeaVW 95% e[WUacWiRQ efficieQc\.
SROXWiRQ
UVe 8 ideaO VWageV
3.21b. Kremser equations: absorption of h\drogen sulfide.
A VchePe fRU Whe UePRYaO Rf H2S fURP a fORZ Rf 1.0 VWd P
3/V Rf QaWXUaO gaV b\ VcUXbbiQg ZiWh
ZaWeU aW 298 K aQd 10 aWP iV beiQg cRQVideUed. The iQiWiaO cRPSRViWiRQ Rf Whe feed gaV iV 2.5 PROe
SeUceQW H2S. A fiQaO gaV VWUeaP cRQWaiQiQg RQO\ 0.1 PROe SeUceQW H2S iV deViUed. The abVRUbiQg
ZaWeU ZiOO eQWeU Whe V\VWeP fUee Rf H2S. AW Whe giYeQ WePSeUaWXUe aQd SUeVVXUe, Whe V\VWeP ZiOO
fROORZ HeQU\'V OaZ, accRUdiQg WR Yi = 48.3Xi, ZheUe Xi = PROeV H2S/PROe Rf ZaWeU; Yi = PROeV
H2S/PROe Rf aiU. 
a) FRU a cRXQWeUcXUUeQW abVRUbeU, deWeUPiQe Whe fORZ UaWe Rf ZaWeU WhaW iV UeTXiUed if 1.5 WiPeV Whe
minimum floZ 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 Zith chemical reaction: H2S scrubbing ZithMEA.
AV VhRZQ iQ PURbOeP 3-21, VcUXbbiQg Rf h\dURgeQ VXOfide fURP QaWXUaO gaV XViQg ZaWeU iV QRW
SUacWicaO ViQce iW UeTXiUeV OaUge aPRXQWV Rf ZaWeU dXe WR Whe ORZ VROXbiOiW\ Rf H2S iQ ZaWeU. If a 2N
VROXWiRQ Rf PRQReWhaQROaPiQe (MEA) iQ ZaWeU iV XVed aV Whe abVRUbeQW, hRZeYeU, Whe UeTXiUed
OiTXid fORZ UaWe iV UedXced dUaPaWicaOO\ becaXVe Whe MEA UeacWV ZiWh Whe abVRUbed H2S iQ Whe
OiTXid ShaVe, effecWiYeO\ iQcUeaViQg iWV VROXbiOiW\.
FRU WhiV VROXWiRQ VWUeQgWh aQd a WePSeUaWXUe Rf 298 K, Whe VROXbiOiW\ Rf H2S caQ be aSSUR[iPaWed
b\ (de NeYeUV, N., Air PollXWion ConWrol Engineering, 2Qd ed., McGUaZ-HiOO, BRVWRQ, MA, 2000):
ReSeaW Whe caOcXOaWiRQV Rf PURbOeP 3.21, bXW XViQg a 2N PRQReWhaQROaPiQe VROXWiRQ aV
abVRUbeQW.
SROXWiRQ
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., PUoceVV EngineeUing and DeVign foU AiU PollXWion ConWUol,
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
3.24b. Kremser equations: absorption of sulfur dio[ide.
An abVoUbeU iV aYailable Wo WUeaW Whe flXe gaV of PUoblem 3.23 Zhich iV eTXiYalenW Wo 8.5
eTXilibUiXm VWageV. 
a) CalcXlaWe Whe ZaWeU floZ UaWe Wo be XVed in WhiV abVoUbeU if 90% of Whe SO2 iV Wo be UemoYed.
CalcXlaWe alVo Whe SO2 concenWUaWion in Whe ZaWeU leaYing Whe abVoUbeU.
SolXWion
IniWial eVWimaWe
b) WhaW iV Whe SeUceQWage UePRYaO Rf SO2 WhaW caQ be achieYed ZiWh WhiV abVRUbeU if Whe ZaWeU fORZ
UaWe XVed iV Whe VaPe WhaW ZaV caOcXOaWed iQ PURbOeP 3.23 (a)?
SROXWiRQ
IQiWiaO eVWiPaWe
3.25b. Kremser equations: liquid e[traction.
AQ aTXeRXV aceWic acid VROXWiRQ fORZV aW Whe UaWe Rf 1,000 Ng/hU. The VROXWiRQ iV 1.1% (b\ ZeighW)
aceWic acid. IW iV deViUed WR UedXce Whe cRQceQWUaWiRQ Rf WhiV VROXWiRQ WR 0.037% (b\ ZeighW) aceWic
acid b\ e[WUacWiRQ ZiWh 3-heSWaQRO aW 298 K. The iQOeW 3-heSWaQRO cRQWaiQV 0.02% (b\ ZeighW) aceWic
acid. AQ e[WUacWiRQ cROXPQ iV aYaiOabOe Zhich iV eTXiYaOeQW WR a cRXQWeUcXUUeQW caVcade Rf 15
eTXiOibUiXP VWageV. WhaW VROYeQW fORZ UaWe iV UeTXiUed? CaOcXOaWe Whe cRPSRViWiRQ Rf Whe VROYeQW
ShaVe OeaYiQg Whe cROXPQ. FRU WhiV V\VWeP, eTXiOibUiXP iV giYeQ b\
WW UaWiR aceWic acid iQ VROYeQW = 0.828 u WW UaWiR aceWic acid iQ ZaWeU
SROXWiRQ
LeW Whe aTXeRXV ShaVe be Whe V-ShaVe; Whe VROYeQW ShaVe iV Whe L-ShaVe.
Initial estimate:
3.26c. CoXnWeUcXUUenW YeUVXV cUoVV-floZ e[WUacWion.
A 1-butanol acid solution is to be e[tracted Zith pure Zater. The butanol solution contains
4.5% (b\ Zeight) of acetic acid and floZs at the rate of 400 kg/hr. A total Zater floZ rate of
1005 kg/hr is used. Operation is at 298 K and 1 atm. For practical purposes, 1-butanol and
Zater are inmiscible. At 298 K, the equilibrium data can be represented b\ YAi = 0.62 XAi,
Zhere YAi is the Zeight ratio of acid in the aqueous phase and XAi is the Zeight ratio of acid in
the organic phase.
a) If the outlet butanol stream is to contain 0.10% (b\ Zeight) acid, hoZ man\ equilibrium
stages are required for a countercurrent cascade?
Solution
b) If Whe ZaWeU iV VSOiW XS eTXaOO\ aPRQg Whe VaPe QXPbeU Rf VWageV, bXW iQ a cURVV-fORZ caVcade,
ZhaW iV Whe RXWOeW 1-bXWaQRO cRQceQWUaWiRQ (Vee PURbOeP 3.19)?
SROXWiRQ
3.27c. Glucose sorption on an ion e[change resin.
ChiQg aQd RXWhYeQ (AIChE S\mp. Ser., 81,S. 242, 1985) fRXQd WhaW Whe eTXiOibUiXP Rf gOXcRVe RQ aQ
iRQ e[chaQge UeViQ iQ Whe caOciXP fRUP ZaV OiQeaU fRU cRQceQWUaWiRQV beORZ 50 g/L. TheiU eTXiOibUiXP
e[SUeVViRQ aW 303 K iV YAi = 1.961 XAi, ZheUe XAi iV Whe gOXcRVe cRQceQWUaWiRQ iQ Whe UeViQ.(g Rf
gOXcRVe SeU OiWeU Rf UeViQ) aQd YAi iV Whe gOXcRVe cRQceQWUaWiRQ iQ VROXWiRQ.(g Rf gOXcRVe SeU OiWeU Rf
VROXWiRQ). 
a) We ZiVh WR VRUb gOXcRVe RQWR WhiV iRQ e[chaQge UeViQ aW 303 K iQ a cRXQWeUcXUUeQW caVcade Rf
ideaO VWageV. The cRQceQWUaWiRQ Rf Whe feed VROXWiRQ iV 15 g/L. We ZaQW aQ RXWOeW cRQceQWUaWiRQ Rf 1.0
g/L. The iQOeW UeViQ cRQWaiQV 0.25 g Rf gOXcRVe/L. The feed VROXWiRQ fORZV aW Whe UaWe Rf 100 L/PiQ,
ZhiOe Whe UeViQ fORZV aW Whe UaWe Rf 250 L/PiQ. FiQd Whe QXPbeU Rf eTXiOibUiXP VWageV UeTXiUed.
SROXWiRQ
b) If 5 equilibrium stages are added to the cascade of part a), calculate the resin floZ required to
maintain the same degree of glucose sorption.
Solution
4.1a. Void fraction near the Zalls of packed beds.
CRQVideU a c\OiQdUicaO YeVVeO ZiWh a diaPeWeU Rf 305 PP SacNed ZiWh VROid VSheUeV ZiWh a diaPeWeU
Rf 50 PP.
a) FURP eTXaWiRQ (4-1), caOcXOaWe Whe aV\PSWRWic SRURViW\ Rf Whe bed.
SROXWiRQ
AQVZeU
b) EVWiPaWe Whe YRid fUacWiRQ aW a diVWaQce Rf 100 PP fURP Whe ZaOO.
AQVZeU
4.2b. Void fraction near the Zalls of packed beds.
BecaXVe Rf Whe RVciOOaWRU\ QaWXUe Rf Whe YRid-fUacWiRQ UadiaO YaUiaWiRQ Rf SacNed bedV, WheUe aUe a
QXPbeU Rf ORcaWiRQV cORVe WR Whe ZaOO ZheUe Whe ORcaO YRid fUacWiRQ iV e[acWO\ eTXaO WR Whe aV\PSWRWic
YaOXe fRU Whe bed. FRU Whe bed deVcUibed iQ E[aPSOe 4.1, caOcXOaWe Whe diVWaQce fURP Whe ZaOO WR Whe
fiUVW fiYe VXch ORcaWiRQV.
SROXWiRQ
The ORcaO SRURViW\ QeaU Whe ZaOO eTXaOV Whe aV\PSWRWic SRURViW\ ZheQ J0(DUd) = 0. FURP
E[aPSOe 4.1,
IQiWiaO eVWiPaWeV Rf Whe URRWV caQ be RbWaiQed fURP Fig. 4.4
IQiWiaO eVWiPaWe
IQiWiaO eVWiPaWe
IQiWiaO eVWiPaWe
IQiWiaO eVWiPaWe
IQiWiaO eVWiPaWe
4.3c. Void fraction near the Zalls of packed beds.
(a) ShRZ WhaW Whe UadiaO ORcaWiRQ Rf Whe Pa[iPa aQd PiQiPa Rf Whe fXQcWiRQ deVcUibed b\ ETXaWiRQ
(4-1) aUe Whe URRWV Rf Whe eTXaWiRQ
SROXWiRQ
b) FRU Whe SacNed bed Rf E[aPSOe 4.1, caOcXOaWe Whe UadiaO ORcaWiRQ Rf Whe fiUVW fiYe Pa[iPa, aQd Rf
Whe fiUVW fiYe PiQiPa; caOcXOaWe Whe aPSOiWXde Rf Whe YRid fUacWiRQ RVciOaWiRQV aW WhRVe SRiQWV.
SROXWiRQ
IniWial eVWimaWeV of Whe UooWV can be obWained fUom Fig. 4.4
IniWial eVWimaWe
IniWial eVWimaWe
RepeaWing WhiV pUocedXUe, Whe folloZing UeVXlWV aUe obWained:
Ma[ima: Minima
U, mm (U*) AmpliWXde, % U,mm (U*) AmpliWXde, %
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) CalcXlaWe Whe diVWance fUom Whe Zall aW Zhich Whe abVolXWe YalXe of Whe poUoViW\ flXcWXaWionV haV
been dampened Wo leVV Whan 10% of Whe aV\mpWoWic bed poUoViW\.
SolXWion
FUom Whe UeVXlWV of paUW (b), WhiV mXVW happen aW U* beWZeen 3.39 and 3.85
(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 HaY/Hb.
Solution
4.4c. Void fraction near the Zalls 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
cylindrical containers of outside diameterDo, inside diameter Di, equivalent diameter De = Do ç Di,
consisting of equal-sized spheres of diameter dp, with diameter aspect ratios of 4 ð De/dp ð 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
(b) Plot the void fraction, as predicted by Eq. (4-66), for r* from 0 to R*.
(c) ShoZ WhaW Whe aYeUage poUoViW\ foU an APB iV giYen b\
(d) EVWimaWe Whe aYeUage poUoViW\ foU Whe APB deVcUibed aboYe.
SolXWion
4.5a. Minimum liquid mass velocit\ for proper wetting of packing.
A 1.0-m diameWeU bed XVed foU abVoUpWion of ammonia ZiWh pXUe ZaWeU aW 298 K iV packed ZiWh
25-mm plaVWic InWalo[ VaddleV. CalcXlaWe Whe minimXm ZaWeU floZ UaWe, in kg/V, neded Wo enVXUe
pUopeU ZeWWing of Whe packing VXUface.
SolXWion
FoU plaVWic packing, YL,min = 1.2 mm/V.
FUom SWeam TableV
4.6a. Minimum liquid mass velocit\ for proper wetting of packing.
RepeaW PUoblem 4.5, bXW XVing ceUamic inVWead of plaVWic InWalo[ VaddleV.
SolXWion
FoU ceUamic packing, YL,min = 0.15 mm/V.
From Steam Tables
4.7b. SSecific OiTXid hROdXS aQd YRid fUacWiRQ iQ fiUVW-geQeUaWiRQ UaQdRP
SacNiQg.
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. SSecific OiTXid hROdXS aQd YRid fUacWiRQ iQ VWUXcWXUed SacNiQg.
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 mç1, H = 0.979, Ch = 0.547 (Seader and Henley,
1998).
Solution
From Example 4.2:
4.9b. SSecific OiTXid hROdXS aQd YRid fUacWiRQ iQ fiUVW-geQeUaWiRQ UaQdRP SacNiQg.
A WRZeU SacNed ZiWh 25-PP ceUaPic RaVchig UiQgV iV WR be XVed fRU abVRUbiQg beQ]eQe YaSRU fURP a
diOXWe Pi[WXUe ZiWh aQ iQeUW gaV XViQg a ZaVh RiO aW 300 K. The YiVcRViW\ Rf Whe RiO iV 2.0 cP aQd iWV
deQViW\ iV 840 Ng/P3. The OiTXid PaVV YeORciW\ iV L' = 2.71 Ng/P2-V. EVWiPaWe Whe OiTXid hROdXS, Whe
YRid fUacWiRQ, aQd Whe h\dUaXOic VSecific aUea Rf Whe SacNiQg.
SROXWiRQ
FURP TabOe 4.1:
4.10b. Pressure drop in beds packed Zith first-generation random packings.
Repeat E[ample 4.3, but using 15-mm ceramic Raschig rings as packing material. Assume that, for
this packing, CS = 1.783.
Solution
Packed ColXmn Design Program
This program calculates the diameter of a packed
column to satisf\ a given pressure drop criterium,
and estimates the volumetric mass-transfer coefficients.
Enter data related to the gas and liquid streams
Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s
Enter liquid densit\, in kg/m3 Enter gas densit\, kg/m3
Enter liquid viscosit\, Pa-s Enter gas viscosit\, Pa-s
Enter temperature, T, in K Enter total pressure, P, in Pa
Enter data related to the packing
Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3
Introduce a units conversion factor in Fp
Enter porosit\, fraction Enter loading constant, Ch
Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m
Calculate flow parameter, X
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
Calculate effective particle si]e, dp, in m
IWeUaWe Wo find Whe WoZeU diameWeU foU Whe giYen pUeVVXUe dUop
ColXmn diameWeU, in meWeUV
FUacWional appUoach Wo flooding
4.11b. PUeVVXUe dUop and appUoach Wo flooding in VWUXcWXUed 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 ft
2/ft3, a = 200 mç1, H = 0.979, Ch = 0.547, Cp =
0.355 (Seader and Henley, 1998).
Solution
Packed ColXmn Design Program
This program calculates the diameter of a packed
column to satisf\ a given pressure drop criterium,
and estimates the volumetric mass-transfer coefficients.
Enter data related to the gas and liquid streams
Enter liquid flow rate, mL, in kg/s Enter gas flow rate, mG, in kg/s
Enter liquid densit\, in kg/m3 Enter gas densit\, kg/m3
Enter liquid viscosit\, Pa-s Enter gas viscosit\, Pa-s
Enter temperature, T, in K Enter total pressure, P, in Pa
Enter data related to the packing
Enter packing factor, Fp, in ft2/ft3 Enter specific area, a, m2/m3
Introduce a units conversion factor in Fp
Enter porosit\, fraction Enter loading constant, Ch
Enter pressure drop constant, Cp Enter allowed pressure drop, in Pa/m
Calculate flow parameter, X
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
Calculate effective particle si]e, dp, in m
IWeUaWe Wo find Whe WoZeU diameWeU foU Whe giYen pUeVVXUe dUop
ColXmn diameWeU, in meWeUV
FUacWional appUoach Wo flooding
4.12b, d. Pressure drop in beds packed Zith second-generation random packings.
A packed toZer is to be designed for the countercurrent contact of a ben]ene-nitrogen gas mi[ture
Zith kerosene to Zash out the ben]ene from the gas. The gas enters the toZer 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
abVRUbed b\ Whe NeURVeQe. The OiTXid eQWeUV Whe WRZeU aW Whe UaWe Rf 4.0 Ng/V; Whe OiTXid deQViW\ iV
800 Ng/P3, YiVcRViW\ iV 2.3 cP. The SacNiQg ZiOO be 50-PP PeWaO PaOO UiQgV, aQd Whe WRZeU diaPeWeU
ZiOO be chRVeQ WR SURdXce a gaV-SUeVVXUe dURS Rf 400 Pa/P Rf iUUigaWed SacNiQg.
(a) CaOcXOaWe Whe WRZeU diaPeWeU WR be XVed, aQd Whe UeVXOWiQg fUacWiRQaO aSSURach WR fORRdiQg.
(b) AVVXPe WhaW, fRU Whe diaPeWeU chRVeQ, Whe iUUigaWed SacNed heighW ZiOO be 5 P aQd WhaW 1 P Rf
XQiUUigaWed SacNiQg ZiOO be SOaced RYeU Whe OiTXid iQOeW WR acW aV eQWUaiQPeQW VeSaUaWRU. The bORZeU-
PRWRU cRPbiQaWiRQ WR be XVed aW Whe gaV iQOeW ZiOO haYe aQ RYeUaOO PechaQicaO efficieQc\ Rf 60%.
CaOcXOaWe Whe SRZeU UeTXiUed WR bORZ Whe gaV WhURXgh Whe SacNiQg.
SROXWiRQ
(a) DeVigQ fRU cRQdiWiRQV aW Whe bRWWRP Rf Whe WRZeU ZheUe Whe Pa[iPXP fORZ Rf gaV aQd OiTXid RccXU
BeQ]eQe eQWeUiQg ZiWh Whe gaV:
AVVXPiQg WhaW aOO Rf Whe beQ]eQe iV abVRUbed:
FURP Whe LXcaV PeWhRd fRU Pi[WXUeV Rf gaVeV:
UViQg Whe PacNed CROXPQ DeVigQ PURgUaP:
D = 0.913 P f = 0.825
(b) CaOcXOaWe Whe SUeVVXUe dURS WhURXgh Whe dU\ SacNiQg RQ WRS
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 u 10ç10 m2/s.
From the wilke-Lee equation, DG = 0.0885 cm
2/s
From Table 4.2, CL = 1.192, CV =
0.410Using the Packed Column Design Program:
kLah = 0.00675 s
ç1; kyah = 0.26 kmol/m
3-s
4.13b, d. Pressure drop in beds packed Zith 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 ft
2/ft3, a = 200 mç1, 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
ç1; kyah = 0.376 kmol/m
3-s
4.14c, d. Air stripping of ZasteZater in a packed column.
A wastewater stream of 0.038 m3/s, containing 10 ppm (by weight) of benzene, is to be stripped
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 lawconstant 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 u 10ç5 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
Using the Packed Column Design Program:
D = 1.145 m f = 0.766 kLah = 0.032 s
ç1; kyah = 0.335 kmol/m
3-s
4.15b. Stripping chloroform from water b\ sparging with air.
Repeat E[ample 4.5, but using an air floZ rate that is tZice the minimum required.
Solution
Initial estimate of the column height, Z
IniWial eVWimaWe of gaV holdXp
Calculate poZer required
4.16b. Stripping chloroform from water b\ sparging with air.
Repeat Example 4.5, but using the same air flow rate used in Problem 4.15, and specifying a
chloroform removal efficiency of 99%. 
Solution
For 99% removal efficiency, xin/xout = 100
Z = 1.30 m
WT(Z) = 5.0 kW
4.17b. Stripping chlorine from water b\ sparging with air.
A vessel 2.0 m in diameter and 2.0 m deep (measured from the gas sparger at the bottom to liquid
overflow at the top) is to be used for stripping chlorine from water by sparging with air. The water
will flow continuously downward at the rate of 7.5 kg/s with an initial chlorine concentration of 5
mg/L. Airflow will be 0.22 kg/s at 298 K. The sparger is in the form of a ring, 25 cm in diameter,
containing 200 orifices, each 3.0 mm in diameter. Henry's law constant for chlorine in water at this
temperature is 0.11 kPa-m3/mole (Perry and Chilton, 1973). The diffusivity of chlorine at infinite
dilution in water at 298 K is 1.25 u 10ç5 cm2/s (Cussler, 1997).
(a) Assuming that all the resistance to mass transfer resides in the liquid phase, estimate the
chlorine removal efficiency achieved.
Solution
IniWial eVWimaWe of gaV holdXp
IWeUaWiQg XQWiO Z = 2.0 P:
(b) CaOcXOaWe SRZeU UeTXiUed
4.18c,d. Batch ZasteZater aeration using spargers.
IQ Whe WUeaWPeQW Rf ZaVWeZaWeU, XQdeViUabOe gaVeV aUe fUeTXeQWO\ VWUiSSed RU deVRUbed fURP Whe
ZaWeU, aQd R[\geQ iV adVRUbed iQWR Whe ZaWeU ZheQ bXbbOeV Rf aiU aUe diVSeUVed QeaU Whe bRWWRP Rf
aeUaWiRQ WaQNV RU SRQdV. AV Whe bXbbOeV UiVe, VROXWe caQ be WUaQVfeUUed fURP Whe gaV WR Whe OiTXid RU
fURP Whe OiTXid WR Whe gaV deSeQdiQg XSRQ Whe cRQceQWUaWiRQ dUiYiQg fRUce. 
FRU baWch aeUaWiRQ iQ a cRQVWaQW YROXPe WaQN, aQ R[\geQ PaVV-baOaQce caQ be ZUiWWeQ aV
ZheUe cA* iV Whe R[\geQ VaWXUaWiRQ cRQceQWUaWiRQ. IQWegUaWiQg beWZeeQ Whe WiPe OiPiWV ]eUR aQd W
aQd Whe cRUUeVSRQdiQg diVVROYed R[\geQ cRQceQWUaWiRQ OiPiWV cA,0 aQd cA,W; aVVXPiQg WhaW cA*
UePaiQV eVVeQWiaOO\ cRQVWaQW, aQd WhaW aOO Whe UeViVWaQce WR PaVV WUaQVfeU UeVideV iQ Whe OiTXid
ShaVe:
IQ aeUaWiRQ WaQNV, ZheUe aiU iV UeOeaVed aW aQ iQcUeaVed OiTXid deSWh, Whe VROXbiOiW\ Rf R[\geQ iV
iQfOXeQced bRWh b\ Whe iQcUeaViQg SUeVVXUe Rf Whe aiU eQWeUiQg Whe aeUaWiRQ WaQN aQd b\ Whe
decUeaViQg R[\geQ SaUWiaO SUeVVXUe iQ Whe aiU bXbbOe aV R[\geQ iV abVRUbed. FRU WheVe caVeV, Whe
XVe Rf a PeaQ VaWXUaWiRQ YaOXe cRUUeVSRQdiQg WR Whe aeUaWiRQ WaQN PiddeSWh iV VXggeVWed
(EcNeQfeOdeU, W. W. JU., IndXVWUial WaWeU PollXWion ConWUol, 3Ud ed., McGUaZ-HiOO, BRVWRQ, Ma,
2000):
ZheUe 
cV = VaWXUaWiRQ diVVROYed R[\geQ cRQceQWUaWiRQ iQ fUeVh ZaWeU e[SRVed
WR aWPRVSheUic aiU aW 101.3 NPa cRQWaiQiQg 20.9% R[\geQ,
PR = abVROXWe SUeVVXUe aW Whe deSWh Rf aiU UeOeaVe,
PV = aWPRVSheUic SUeVVXUe,
OW = PROaU R[\geQ SeUceQW iQ Whe aiU OeaYiQg Whe aeUaWiRQ WaQN.
The PROaU R[\geQ SeUceQW iQ Whe aiU OeaYiQg Whe aeUaWiRQ WaQN iV UeOaWed WR Whe R[\geQ WUaQVfeU
efficieQc\, Oeff, WhURXgh 
ZheUe
CRQVideU a 567 P3 aeUaWiRQ SRQd aeUaWed ZiWh 15 VSaUgeUV, each XViQg cRPSUeVVed aiU aW a UaWe Rf
0.01 Ng/V. Each VSaUgeU iV iQ Whe fRUP Rf a UiQg, 100 cP iQ diaPeWeU, cRQWaiQiQg 20 RUificeV, each
3.0 PP iQ diaPeWeU. The VSaUgeUV ZiOO be ORcaWed 5 P beORZ Whe VXUface Rf Whe SRQd. The ZaWeU
WePSeUaWXUe iV 298 K; aWPRVSheUic cRQdiWiRQV aUe 298 K aQd 101.3 NPa. UQdeU WheVe cRQdiWiRQV,
cV = 8.38 Pg/L (DaYiV aQd CRUQZeOO, 1998).
(a) EVWiPaWe Whe YROXPeWUic PaVV-WUaQVfeU cRefficieQW fRU WheVe cRQdiWiRQV fURP eTXaWiRQV (4-23)
aQd (4-25).
(b) EVWiPaWe Whe WiPe UeTXiUed WR UaiVe Whe diVVROYed R[\geQ cRQceQWUaWiRQ fURP 0.5 Pg/L WR 6.0
Pg/L, aQd caOcXOaWe Whe UeVXOWiQg R[\geQ WUaQVfeU efficieQc\.
(c) EVWiPaWe Whe SRZeU UeTXiUed WR RSeUaWe Whe 15 VSaUgeUV, if Whe PechaQicaO efficieQc\ Rf Whe
cRPSUeVVRU iV 60%.
SROXWiRQ
IniWial eVWimaWe of gaV holdXp
Initial estimate of transfer efficienc\
(b) Calculate power required
4.19c,d. Batch ZasteZater aeration using spargers; effect of liquid depth.
CRQVideU Whe ViWXaWiRQ deVcUibed iQ PURbOeP 4.18. AccRUdiQg WR EcNeQfeOdeU (2000), fRU PRVW W\SeV
Rf bXbbOe-diffXViRQ aeUaWiRQ V\VWePV Whe YROXPeWUic PaVV-WUaQVfeU cRefficieQW ZiOO YaU\ ZiWh OiTXid
deSWh Z accRUdiQg WR Whe UeOaWiRQVhiS
ZheUe Whe e[SRQeQW Q haV a YaOXe QeaU 0.7 fRU PRVW V\VWePV. FRU Whe aeUaWiRQ SRQd Rf PURbOeP
4.18, caOcXOaWe NLa aW YaOXeV Rf Z = 3P, 4P, 6P, aQd 7 P. EVWiPaWe Whe cRUUeVSRQdiQg YaOXe Rf Q
fURP UegUeVViRQ aQaO\ViV Rf Whe UeVXOWV. HiQW: RePePbeU WhaW Whe WRWaO YROXPe Rf Whe SRQd PXVW
UePaiQ cRQVWaQW, WheUefRUe Whe cURVV-VecWiRQaO aUea Rf Whe SRQd PXVW chaQge aV Whe ZaWeU deSWh
chaQgeV.
SROXWiRQ
UViQg Whe SURgUaP deYeORSed iQ PURb 4.18, Whe fROORZiQg YecWRU Rf UeVXOWV iV geQeUaWed
4.20c. Flooding conditions in a packed cooling toZer.
A cRROiQg WRZeU, 2 P iQ diaPeWeU, SacNed ZiWh 75-PP ceUaPic HifORZ UiQgV, iV fed ZiWh ZaWeU aW 316
K aW a UaWe Rf 25 Ng/P2-V. The ZaWeU iV cRQWacWed ZiWh aiU, aW 300 K aQd 101.3 NPa eVVeQWiaOO\ dU\,
dUaZQ XSZaUd cRXQWeUcXUUeQWO\ WR Whe ZaWeU fORZ. NegOecWiQg eYaSRUaWiRQ Rf Whe ZaWeU aQd chaQgeV
iQ Whe aiU WePSeUaWXUe, eVWiPaWe Whe YROXPeWUic UaWe Rf aiUfORZ, iQ P3/V, Zhich ZRXOd fORRd Whe WRZeU.
SROXWiRQ
LMV = OiTXid PaVV YeORciW\ GMV = gaVV PaVV YeORciW\
InWUodXce a XniWV conYeUVion facWoU in Fp
IniWial eVWimaWe of GMV
4.21c,d. Design of a sieve-tra\ column for ethanol absorption
RepeaW Whe calcXlaWionV of E[ampleV 4.6, 4.7, 4.8, and 4.9 foU a colXmn diameWeU coUUeVponding Wo
50% of flooding.
SolXWion
FUom Whe SieYe-PlaWe DeVign PUogUam, Whe folloZing UeVXlWV aUe obWained foU I = 0.5
D = 1.176 m W = 0.6 m 'P = 590 Pa/WUa\
FUoXde No. = 1.22 (no e[ceVViYe Zeeping) E = 0.0165
EOG = 0.8123 EMG = 0.8865 EMGE = 0.878
4.22c,d. Design of a sieve-tra\ column for aniline stripping.
A sieve-tray tower is to be designed for stripping an aniline (C6H7N)-water solution with steam.
The circumstances at the top of the tower, which are to be used to establish the design, are:
Temperature = 371.5 K Pressure = 100 kPa
Liquid:
Rate = 10.0 kg/s Composition = 7.00 mass % aniline
Density = 961 kg/m3 Viscosity = 0.3 cP
Surface tension = 58 dyne/cm 
Diffusivity = 4.27 u 10ç5 cm2/s (est.) Foaming factor = 0.90
Vapor:
Rate = 5.0 kg/s Composition = 3.6 mole % aniline
Density = 0.670 kg/m3 Viscosity = 118 mP (est.)
Diffusivity = 0.116 cm2/s (est.)
The equilibrium data at this concentration indicates that P = 0.0636 (Treybal, 1980). 
(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 5.5 mm on an equilateral-
triangular pitch 12 mm between hole centers, punched in stainless steel sheet metal 2 mm
thick. Use a weir height of 40 mm. Design for a 75% approach to the flood velocity. Report
details respecting tower diameter, tray spacing, weir length, gas-pressuredrop, and entrainment
in the gas. Check for excessive weeping.
Solution
For aniline, the molecular weight = 93. The average molecular weight of the gas = 20.7.
The composition of the liquid is calculated as 1.4 mole % aniline, the average molecular weight
= 19.1.
From the Sieve-Plate Design Program, the following results are obtained for f = 0.75
D = 1.93 m t = 0.6 m DP = 435 Pa/tray
Froude No. = 0.93 (no excessive weeping) E = 0.045
Weir length = 1.40 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.716 EMG = 0.722 EMGE = 0.720
4.23c,d. Design of a sieve-tra\ column for aniline stripping.
Repeat Problem 4.22, but for a 45% approach to flooding. Everything else remains the same as in
Problem 4.22.
Solution
From the Sieve-Plate Design Program, the following results are obtained for f = 0.45
D = 2.49 m t = 0.6 m DP = 399 Pa/tray
Froude No. = 0.491 ( excessive weeping) E = 0.0173
Weir length = 1.81 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.660 EMG = 0.666 EMGE = 0.665
4.24c,d. Design of a sieve-tra\ column for methanol stripping.
A dilute aqueous solution of methanol is to be stripped with steam in a sieve-tray tower. The
conditions chosen for design are:
Temperature = 368 K Pressure = 101.3 kPa
Liquid:
Rate = 0.25 kmol/s Composition = 15.0 mass % methanol
Density = 961 kg/m3 Viscosity = 0.3 cP
Surface tension = 40 dyn/cm 
Diffusivity = 5.70 u 10ç5 cm2/s (est.) Foaming factor = 1.0
Vapor:
Rate = 0.1 kmol/s Composition = 18 mole % methanol
Viscosity = 125 mP (est.) Diffusivity = 0.213 cm2/s (est.)
The equilibrium data at this concentration indicates that P = 2.5 (Perry and Chilton, 1973). 
(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 6.0 mm on an equilateral-
triangular pitch 12 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.
Use a weir height of 50 mm. Design for 80% approach to the flood velocity. Report details
respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in the
gas. Check for excessive weeping.
Solution
From the Sieve-Plate Design Program, the following results are obtained for f = 0.80
D = 1.174 m t = 0.6 m DP = 386 Pa/tray
Froude No. = 0.878 (no excessive weeping) E = 0.052
Weir length = 0.853 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.5987 EMG = 0.707 EMGE = 0.700
4.25c,d. Sieve-tra\ column for methanol stripping; effect of hole si]e.
Repeat Problem 4.24, but changing the perforation size to 4.5 mm, keeping everything else
constant.
Solution
From the Sieve-Plate Design Program, the following results are obtained for I = 0.80
D = 1.174 m t = 0.6 m DP = 686 Pa/tray
Froude No. = 1.561 (no excessive weeping) E = 0.076
Weir length = 0.853 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.793 EMG = 0.936 EMGE = 0.918
4.26c,d. Design of sieve-tra\ column for butane absorption.
A gas containing methane, propane, and n-butane is to be scrubbed countercurrently in a sieve-tray
tower with a hydrocarbon oil to absorb principally the butane. It is agreed to design a tray for the
circumstances existing at the bottom of the tower , where the conditions are:
Temperature = 310 K Pressure = 350 kPa
Liquid:
Rate = 0.50 kmol/s Average molecular weight = 150
Density = 850 kg/m3 Viscosity = 1.6 cP
Surface tension = 25 dyn/cm 
Diffusivity = 1.14 u 10ç5 cm2/s (est.) Foaming factor = 0.9
Vapor:
Rate = 0.3 kmol/s Composition = 86% CH4, 12% C3H8, 2% C4H10
Viscosity = 113 mP (est.) Diffusivity = 0.035 cm2/s (est.)
The system obeys Raoult's law; the vapor pressure of n-butane at 310 K is 3.472 bar (Reid et al.,
1987).
(a) Design a suitable cross-flow sieve-tray for such a tower. According to Bennett and Kovak (2000),
the optimal value of the ratio Ah/Aa is that which yields an orifice Froude number, Fro = 0.5. Design
for the optimal value of do on an equilateral-triangular pitch 12 mm between hole centers, punched in
stainless steel sheet metal 2 mm thick. Use a weir height of 50 mm. Design for a 75% approach to
the flood velocity. Report details respecting tower diameter, tray spacing, weir length, gas-pressure
drop, and entrainment in the gas.
Solution
The slope of the equilibrium curve is approximately 1.0. The average molecular weight of the gas is
20.2. By trial-and-error, using the Sieve-Plate Design Program, it was found that the orifice
diameter that results in a Froude number = 0.5 is do = 3.43 mm.
From the Sieve-Plate Design Program, the following results are obtained for I = 0.75
D = 2.73 m t = 0.6 m DP = 811 Pa/tray
Froude No. = 0.5 (no excessive weeping) E = 0.013
Weir length = 2.28 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.833 EMG = 1.038 EMGE = 1.034
4.27c,d. Design of sieve-tra\ column for ammonia absorption.
A process for making small amounts of hydrogen by cracking ammonia is being considered, and
residual uncracked ammonia is to be removed from the resulting gas. The gas will consist of H2
and N2 in the molar ratio 3:1, containing 3% NH3 by volume. The ammonia will be removed by
scrubbing the gas countercurrently with pure liquid water in a sieve-tray tower. Conditions at the
bottom of the tower are:
Temperature = 303 K Pressure = 200 kPa
Liquid:
Rate = 6.0 kg/s Average molecular weight = 18
Density = 996 kg/m3 Viscosity = 0.9 cP
Surface tension = 68 dyn/cm 
Diffusivity = 2.42 u 10ç5 cm2/s (est.) Foaming factor = 1.0
Vapor:
Rate = 0.7 kg/s 
Viscosity = 113 mP (est.) Diffusivity = 0.230 cm2/s (est.)
For dilute solutions, NH3-H2O follows Henry's law, and at 303 K the slope of the equilibrium curve
is m = 0.85 (Treybal, 1980).
(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 4.75 mm on an equilateral-
triangular pitch 12.5 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.
Use a weir height of 40 mm. Design for an 80% approach to the flood velocity. Report details
respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in the
gas. Check for excessive weeping.
Solution
From the Sieve-Plate Design Program, the following results are obtained for I = 0.80
D = 0.775 m t = 0.5 m DP = 624 Pa/tray
Froude No. = 1.033 (no excessive weeping) E = 0.048
Weir length = 2.28 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.806 EMG = 0.85 EMGE = 0.847
4.28c,d. Design of sieve-tra\ column for toluene-meth\lc\clohe[ane distillation.
A sieve-tray tower is to be designed for distillation of a mixture of toluene and methylcyclohexane.
The circumstances which are to be used to establish the design, are:
Temperature = 380 K Pressure = 98.8 kPa
Liquid:
Rate = 4.8 mole/s Composition =48.0 mol % toluene
Density = 726 kg/m3 Viscosity = 0.22 cP
Surface tension = 16.9 dyne/cm 
Diffusivity = 7.08 u 10ç5 cm2/s (est.) Foaming factor = 0.80
Vapor:
Rate = 4.54 mole/s Composition =44.6 mole % toluene
Density = 2.986 kg/m3 Viscosity = 337 mP (est.)
Diffusivity = 0.0386 cm2/s (est.)
The equilibrium data at this concentration indicates that m = 1.152. 
(a) Design a suitable cross-flow sieve-tray for such a tower. Take do = 4.8 mm on an equilateral-
triangular pitch 12.7 mm between hole centers, punched in stainless steel sheet metal 2 mm thick.
Use a weir height of 50 mm. Design for a 60% approach to the flood velocity. Report details
respecting tower diameter, tray spacing, weir length, gas-pressure drop, and entrainment in the
gas. Check for excessive weeping.
(b) Estimate thetray efficiency corrected for entrainment for the design reported in part (a).
SolXWion
a) From the Sieve-Plate Design Program, the following results are obtained for I = 0.60
D = 0.6 m t = 0.5 m DP = 295 Pa/tray
Froude No. = 0.585 (no excessive weeping) E = 0.007
Weir length = 0.436 m
(b) Estimate the tray efficiency corrected for entrainment for the design reported in part (a).
Solution
EOG = 0.611 EMG = 0.712 EMGE = 0.710
5.1a. Overall tra\ efficienc\
CRQVideU a WUa\ abVRUbeU ZiWh a cRQVWaQW MXUShUee efficieQc\ EMGE = 0.75, aQd aQ aYeUage
abVRUSWiRQ facWRU A = 1.25.
a) EVWiPaWe Whe RYeUaOO WUa\ efficieQc\.
SROXWiRQ
b) If Whe abVRUbeU UeTXiUeV 5.34 eTXiOibUiXP VWageV, caOcXOaWe Whe QXPbeU Rf UeaO WUa\V.
SROXWiRQ
UVe 8 WUa\V
5.3a. Ammonia stripping from a wastewater in a tra\ tower.
A WUa\ WRZeU SURYidiQg Vi[ eTXiOibUiXP VWageV iV XVed fRU VWUiSSiQg aPPRQia fURP a ZaVWeZaWeU
VWUeaP b\ PeaQV Rf cRXQWeUcXUUeQW aiU aW 1 aWP aQd 300 K. CaOcXOaWe Whe cRQceQWUaWiRQ Rf
aPPRQia iQ Whe e[iW ZaWeU if Whe iQOeW OiTXid cRQceQWUaWiRQ iV 0.1 PROe % aPPRQia, Whe iQOeW aiU iV
fUee Rf aPPRQia, aQd 1.873 VWaQdaUd cXbic PeWeU Rf aiU aUe fed WR Whe WRZeU SeU NiORgUaP Rf
ZaVWeZaWeU. The eTXiOibUiXP daWa fRU WhiV V\VWeP, iQ WhiV UaQge Rf cRQceQWUaWiRQV aQd 300 K, caQ
be UeSUeVeQWed b\ \A,i= 1.414 [A,i (KiQg, C. J., SepaUaWion PUoceVVeV, McGUaZ-HiOO, NeZ YRUN,
NY, 1971). 
SROXWiRQ
Initial estimate
5.4a. Ammonia stripping from a wastewater in a tra\ tower.
The Murphree plate efficienc\ for the ammonia stripper of Problem 5.3 is constant at 0.581.
Estimate the number of real tra\s required.
Solution
Use 9 tra\s
5.5b. Ammonia stripping from a wastewater in a tra\ tower.
If the air flow rate to the absorber of Problems 5.3 and 5.4 is reduced to 1.5 standard m3/kg of
ZaWeU, caOcXOaWe cRQceQWUaWiRQ Rf aPPRQia iQ Whe e[iW ZaWeU if 9 UeaO WUa\V aUe XVed aQd Whe
MXUShUee efficieQc\ UePaiQV cRQVWaQW aW 58.1%.
SROXWiRQ
IQiWiaO eVWiPaWe
5.6b. Absorption of an air pollutant in a tra\ tower.
A heaY\-RiO VWUeaP aW 320 K iV XVed iQ aQ abVRUbeU WR UePRYe diOXWe TXaQWiWieV Rf SROOXWaQW A fURP
aQ aiU VWUeaP. The heaY\ RiO iV WheQ Uec\cOed bacN WR Whe SURceVV ZheUe ZheUe A iV VWUiSSed. The
SURceVV iV beiQg UXQ RQ a SiORW SOaQ baViV, aQd iQfRUPaWiRQ fRU VcaOe-XS iV deViUed. The cXUUeQW
abVRUbeU iV a 16-SOaWe VieYe-WUa\ cROXPQ. PiORW SOaW daWa aUe aV fROORZV:
LiTXid fORZ UaWe = 5.0 PROeV/hU
GaV fORZ UaWe = 2.5 PROeV/hU
\A,iQ = 0.04 \A,RXW = 0.0001 [A,iQ = 0.0001
ETXiOibUiXP fRU A iV giYeQ aV \A,i = 0.7[A,i. FiQd Whe RYeUaOO cROXPQ efficieQc\, aQd Whe MXUShUee
SOaWe efficieQc\. aVVXPe WhaW Whe OiTXid aQd gaV fORZ UaWeV aUe URXghO\ cRQVWaQW.
SROXWiRQ
5.7b. Absorption of ammonia in a laborator\-scale tra\ tower.
An absorption column for laboratory use has been carefully constructed so that it has exactly 4
equilibrium stages, and is being used to measure equilibrium data. Water is used as the solvent
to absorb ammonia from air. The system operates isothermally at 300 K and 1 atm. The inlet
water is pure distilled water. The ratio of L/V = 1.2, inlet gas concentration is 0.01 mole fraction
ammonia; the measured outlet gas concentration is 0.0027 mole fraction ammonia. Assuming
that Henry®s law applies, calculate the slope of the equilibrium line.
Solution
Initial estimate
5.8c,d. Absorption of ammonia in a seive-tra\ tower.
A process for making small amounts of hydrogen by cracking ammonia is being considered
and residual, uncracked ammonia is to be removed from the resulting gas. The gas will consist
of H2 and N2 in the molar ratio 3:1, containing 3% NH3 by volume, at a pressure of 2 bars and
a temperature of 303 K.
There is available a sieve-tray tower, 0.75 m diameter, containing 6 cross-flow trays at 0.5 m
tray spacing. The perforations are 4.75 mm in diameter, arranged in triangular pitch on 12.5
mm centers, punched in sheet metal 2 mm thick. The weir height is 40 mm. Assume
isothermal scrubbing with pure water at 303 K. The water flow rate to be used should not
exceed 50% of the maximum recommended for cross-flow sieve trays which is 0.015 m3/s-m
of tower diameter (Treybal, 1980). The gas flow rate should not exceed 80% of the flooding
value.
a) Estimate the gas flow rate that can be processed in the column under the circumstances
described above.
b) Calculate the concentration of the gas leaving the absorber in part a).
Data:
Liquid density = 996 kg/m3 Surface tension = 0.068 N/m
Foaming factor = 1.0 Liquid diffusivity = 2.42 u 10ç9 m2/s
Gas viscosity = 1.122 u 10ç5 kg/m-s Gas diffusivity = 0.23 cm2/s
Slope of the equilibrium line, P = 0.85
Solution
a)
Run the sieve-tray design program of Appendix E with different values
of gas flow rate until D = 0.75 m at f = 0.8. The results are:
mG = 0.657 kg/s Gas pressure drop = 615 Pa per tray
EMGE = 0.841 FUo = 1.042 E = 0.048
b)
IniWial eVWimaWe
5.9b. Absorption of carbon dio[ide in a bubble-cap tra\ toZer.
A SOaQW PaQXfacWXUiQg dU\ ice ZiOO bXUQ cRNe iQ aiU WR SURdXce a fOXe gaV Zhich, ZheQ cOeaQed aQd
cRROed, ZiOO cRQWaiQ 15% CO2, 6% O2, aQd 79% N2. The gaV ZiOO be bORZQ iQWR a bXbbOe-caS WRZeU
VcUXbbeU aW 1.2 aWP aQd 298 K, WR be VcUXbbed cRXQWeUcXUUeQWO\ ZiWh a 30 ZW% PRQReWhaQROaPiQe
(C2H7ON) aTXeRXV VROXWiRQ eQWeUiQg aW 298 K. The VcUXbbiQg OiTXid, Zhich iV Uec\cOed fURP a
VWUiSSeU, ZiOO cRQWaiQ 0.058 PROe CO2/PRO VROXWiRQ. The gaV OeaYiQg Whe VcUXbbeU iV WR cRQWaiQ 2%
CO2. A OiTXid-WR-gaV UaWiR Rf 1.2 WiPeV Whe PiQiPXP iV VSecified. AVVXPe iVRWheUPaO RSeUaWiRQ. AW
298 K aQd 1.2 aWP, Whe eTXiOibUiXP PROe fUacWiRQ Rf caUbRQ diR[ide RYeU aTXeRXV VROXWiRQV Rf
PRQReWhaQROaPiQe (30 ZW%) iV giYeQ b\
ZheUe [A,i iV Whe PROe fUacWiRQ Rf CO2 iQ Whe OiTXid VROXWiRQ.
a) CaOcXOaWe Whe NiORgUaPV Rf VROXWiRQ eQWeUiQg Whe WRZeU SeU cXbic PeWeU Rf eQWeUiQg gaV.
SROXWiRQ
FRU Whe PiQiPXP OiTXid fORZ UaWe, [1Pa[ ZiOO be iQ eTXiOibUiXP ZiWh \N1
IQiWiaO eVWiPaWe
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
c) The monoethanolamine solution has a viscosit\ of 6.0 cP and a densit\ of 1,012 kg/m3.
Estimate the overall tra\ efficienc\ for the absorber, and the number of real tra\s required.
Seader and Henle\ (1998) proposed the following empirical correlation to estimate the 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
UL = liquid densit\, in kg/m
3
Hint: In this problem, the equilibrium-distribution curve is not a straight line, therefore m is not
constant. Estimate thhe average value of m at liquid concentrations along the operating line
and use theaverage in the correlation given.
Solution
5.11c,d. Absorption of carbon disulfide in a sieve-tra\ tower.
Carbon disulfide, CS2, used as a solvent in a chemical plant, is evaporated from the product in a
dryer into an inert gas (essentially N2) in order to avoid an explosion hazard. The CS2-N2 mixture
is to be scrubbed with an absorbent hydrocarbon oil. The gas will flow at the rate of 0.4 m3/s at
297 K and 1 atm. The partial pressure of CS2 in the original gas is 50 mm Hg, and the CS2
concentration in the outlet gas is not to exceed 0.5%. The oil enters the absorber essentially pure
at a rate 1.5 times the minimum, and solutions of oil and CS2 follow Raoult®s law. Design a
sieve-tray tower for this process. Design for a gas velocity which is 70% of the flooding velocity.
Assuming isothermal operation, determine:
a) Liquid flow rate, kg/s
Data (at 297 K):
Oil average molecular weight = 254 Oil viscosity = 4 cP
Oil density = 810 kg/m3 Surfacetension = 0.030 N/m
Foaming factor = 0.9 CS2 vapor pressure = 346 mm Hg 
Solution
IniWial eVWimaWeV
b) ToZer diameter and plate spacing
Solution
Plate design conditions:
f = 0.7 4.5-mm holes in triangular pitch; 12 mm betZeen centers
Weir height = 50 mm; Plate thickness = 2 mm
From the Mathcad program in Appendi[ E:
D = 0.705 m; t = 0.5 m; Pressure drop = 414 Pa/tra\
Fro = 0.712 E = 0.018 EMGE = 0.588
d) Number of real tra\s required
Solution
Appro[imation, assuming straight equilibrium and operating lines
Graphical solution: Generate operating line in [\ diagram
Graphicall\, the number of ideal stages is also slightl\ over 5
Use 10 tra\s
e) Total gas-pressure drop.
Solution
5.12c. Adiabatic absorption of carbon disulfide in a sieve-tra\ tower.
Determine the number of equilibrium tra\s for the absorber of Problem 5.11, assuming adiabatic
operation. The specific heats are:
Substance J/mole-K
Oil 362.2
CS2, liquid 76.2
N2 29.1
CS2, gas 46.9
the latent heat of vapori]ation of CS2 at 297 K is 27.91 kJ/mole. The vapor pressure of CS2 as a
function of temperature is given b\
Solution
Base for enthalpies: liquid oil, liquid CS2, gaseous N2 at 297 K
Assume T1 = 298.0 K
Appro[imatel\ 12 ideal stages
Gas outlet temperature = 298 K
5.13b. Steam-stripping of ben]ene in a sieve-plate column.
A straw oil used to absorb benzene from coke-oven gas is to be steam-stripped in a sieve-plate
column at atmospheric pressure to recover the disolved benzene, C6H6. Equilibrium conditions at
the operating temperature are approximated by Henry®s law such that, when the oil phase contains
10 mole % benzene, the equilibrium benzene partial pressure above the oil is 5.07 kPa. The oil may
be considered nonvolatile. It enters the stripper containing 8 mole % benzene, 75% of which is to be
removed. The steam leaving contains 3 mole % C6H6.
a) How many theorethical stages are required?
b) How many moles of steam are required per 100 moles of the oil-benzene mixture?
c) If 85% of the benzene is to be recovered with the same steam and oil rates, how many theoretical
stages are required? 
Solution 
a)
b)
KUePVeU eTXaWiRQ
c)
(GUaShicaO cRQVWUXcWiRQ ahead)
Kremser equation
5.15c,d. Absorption of carbon disulfide in a random-packed toZer.
Design a toZer packed Zith 50-mm ceramic HifloZ rings for the carbon disulfide scrubber of
Problem 5.11. Assume isothermal operation and use a liquid rate of 1.5 times the minimum and
a gas-pressure drop not e[ceeding 175 Pa/m of packing. Calculate the toZer diameter, packed
height, and total gas-pressure drop. Assume that Ch for the packing is 1.0.
Solution
From Prob. 5.11, the conditions at the bottom of the absorber are:
From Tables 4.1 and 4.2
Run the Packed ToZer Design Program of Ap. D
Results:
Estimate the packed height; assume dilute solutions
From Prob 5.11:
Calculate HtOG at the bottom of the toZer
Calculate HtOG at the top of the tower
From Prob. 5.11, the conditions at the top of the absorber are:
The mass-tranfer coefficients remain fairl\ constant along the tower
5.16b,d. Absorption of sulfur dio[ide in a random-packed toZer.
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, using a packed tower that
is 0.7 m in diameter. The tower is packed with 35-mm plastic NORPAC rings. The average gas-
pressure drop is 200 Pa/m. Equilibrium is described by HenryÕs law with yi = 8.4xi.. If the
liquid flow is adjusted so that the driving force (y ç y*) is constant, calculate the height of the
packed section. Flue gases usually contain less than 1 mole % of SO2, an air pollutant
regulated by law. Assume that the properties of the liquid are similar to those of pure water,
and that the properties of the flue gases are similar to those of air.
Solution
For the driving force to be constant, the operating line is parallel to the equilibrium line, and
the absorption factor A = 1.0
It was found by trial and error with the Packed Column program that, for a pressure drop of
200 Pa/m and D = 0.7 m:
5.17c. Ben]ene vapor recover\ s\stem.
Benzene vapor in the gaseous effluent of an industrial process is scrubbed with a wash oil in a
countercurrent packed absorber. The resulting benzene-wash-oil solution is then heated to 398
K and stripped in a tray tower, using steam as the stripping medium. The stripped wash oil is
then cooled and recycled to the absorber. Some data relative to the operation follow:
Absorption:
Benzene in entering gas = 1.0 mole %
Operating pressure of absorber = 800 mm Hg
Oil circulation rate = 2 m3/1,000 m3 of gas at STP
Oil specific gravity = 0.88 Molecular weight = 260
Henry®s law constant = 0.095 at 293 K; 
= 0.130 at 300 K
NtOG = 5 transfer units
Stripping:
Pressure = 1 atm Steam at 1 atm, 398 K
Henry®s law constant = 3.08 at 398 K
Number of equilibrium stages = 5
a) In the winter, it is possible to cool the recycled oil to 293 K, at which temperature the
absorber then operates. Under these conditions 72.0 kg of steam is used in the stripper per
1,000 m3 of gas at STP entering the absorber. Calculate the percent benzene recovery in the
winter.
Solution
IQiWiaO eVWiPaWeV:
b) IQ Whe VXPPeU iW iV iPSRVVibOe WR cRRO Whe Uec\cOed ZaVh RiO WR ORZeU WhaQ 300 K ZiWh Whe
aYaiOabOe cRROiQg ZaWeU. AVVXPiQg WhaW Whe abVRUbeU WheQ RSeUaWeV aW 300 K, ZiWh Whe VaPe RiO
aQd VWeaP UaWeV, aQd WhaW NWOG aQd eTXiOibUiXP VWageV UePaiQ Whe VaPe, ZhaW VXPPeU
UecRYeU\ Rf beQ]eQe caQ be e[SecWed?
SROXWiRQ
c) If Whe oil UaWe cannoW be incUeaVed, bXW Whe VWeam UaWe in Whe VXmmeU iV incUeaVed b\ 50%
oYeU Whe ZinWeU YalXe, ZhaW VXmmeU UecoYeU\ of ben]ene can be e[pecWed?
5.19c. AbVoUpWion of geUmaniXm WeWUachloUide XVed foU opWical fibeUV.
Germanium tetrachloride (GeCl4) and silicon tetrachloride (SiCl4) are used in the production of
optical fibers. Both chlorides are oxidized at high temperature and converted to glasslike particles.
However, the GeCl4 oxidation is quite incomplete and it is necessary to scrub the unreacted GeCl4
from its air carrier in a packed column operating at 298 K and 1 atm with a dilute caustic solution.
At these conditions, the dissolved GeCl4 has no vapor pressure and mass transfer is controlled by
the gas phase. Thus, the equilibrium curve is a straight line of zero slope. The entering gas flows at
the rate of 23,850 kg/day of air containing 288 kg/day of GeCl4. The air also contains 540 kg/day of
Cl2, which, when dissolved, also will have no vapor pressure.
It is desired to absorb, at least, 99% of both GeCl4 and Cl2 in an existing 0.75-m-diameter column
that is packed to a height of 3.0 m with 13-mm ceramic Raschig rings. The liquid rate should be
set so that the column operates at 75% of flooding. Because the solutions are very dilute, it can be
assumed that both gases are absorbed independenttly.
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 0.68ç2.0 kg/m2-s (Shulman, H. L., et al., AICKE J., 17, 631, 1971):
where:
ds = equivalent packing diameter (0.01774 m for 
13-mm ceramic Raschig rings)
U = gas density, kg/m3
Gx,Gy mass velocities, kg/m
2-s
For the two diffusing species, take DGeCl4 = 0.06 cm
2/s; DCl2 = 013 cm
2/s. For the packing,
Fp = 580 ft
ç1, e = 0.63.
Determine:
a) Liquid flow rate, in kg/s.
Solution
IniWial eVWimaWe
b) The peUcenW abVoUpWion of GeCl4 and Cl2 baVed on Whe aYailable 3.0 m of packing. 
SolXWion (dilXWe VolXWionV)
5.20c,d. Absorption of carbon disulfide in a structured-packed toZer.
Redesignthe absorber of Problem 5.15 using metal Montz B1-300 structured packing. The
characteristics of this packing are (Seader and Henley, 1998):
Fp = 33 ft
ç1 a = 300 mç1 H = 0.93 Ch = 0.482
Cp = 0.295 CL = 1.165 CV = 0.422
Solution
From the Packed Tower Program in Appendix D:
Estimate the packed height; assume dilute solutions
From Prob 5.11:
Calculate HtOG at the bottom of the toZer
Calculate HtOG at the top of the tower
From Prob. 5.11, the conditions at the top of the absorber are:
The mass-tranfer coefficients remain fairl\ constant along the tower
5.21c,d. Absorption of ammonia in a random-packed toZer.
It is desired to reduce the ammonia content of 0.05 m3/s of an ammonia-air mixture (300 K and 1
atm) from 5.0 to 0.04% by volume by water scrubbing. There is available a 0.3-m-diameter tower
packed with 25-mm ceramic Raschig rings to a depth of 3.5 m. Is the tower satisfactory, and if so,
what water rate should be used? At 300 K, ammonia-water solutions follow Henry®s law up to 5
mole % ammonia in the liquid, with m = 1.414.
Solution
Calculate minimum water flow rate
Try
From Appendi[ D
Tr\
From Appendi[ D
Tr\
From Appendi[ D
Tr\
From Appendi[ D
CXbic VSOiQe iQWeUSROaWiRQ
5.22b,d. Absorption of sulfur dio[ide in a structured-packed toZer.
A WRZeU SacNed ZiWh PeWaO MRQW] B1-300 VWUXcWXUed SacNiQg iV WR be deVigQed WR abVRUb SO2
fURP aiU b\ VcUXbbiQg ZiWh ZaWeU. The eQWeUiQg gaV, aW aQ SO2-fUee fORZ UaWe Rf 37.44 PROeV/P
2-V
Rf bed cURVV-VecWiRQaO aUea, cRQWaiQV 20 PROe % Rf SO2. PXUe ZaWeU eQWeUV aW a fORZ UaWe Rf 1976
PROeV/P2-V Rf bed cURVV-VecWiRQaO aUea. The e[iWiQg gaV iV WR cRQWaiQ RQO\ 0.5 PROe % SO2.
AVVXPe WhaW QeiWheU aiU QRU ZaWeU ZiOO WUaQVfeU beWZeeQ Whe ShaVeV aQd WhaW Whe WRZeU RSeUaWeV
iVRWheUPaOO\ aW 2 aWP aQd 303 K. ETXiOibUiXP daWa fRU VROXbiOiW\ Rf SO2 iQ ZaWeU aW 303 N aQd 1
aWP haYe beeQ fiWWed b\ OeaVW-VTXaUeV WR Whe eTXaWiRQ (SeadeU aQd HeQOe\, 1998):
a) DeUiYe Whe fROORZiQg RSeUaWiQg OiQe eTXaWiRQ fRU Whe abVRUbeU:
SROXWiRQ
b) If the absorber is to process 1.0 m3/s (at 2 atm and 303 K) of the entering gas, calculate
the water flow rate, the tower diameter, and the gas-pressure drop per unit of packing height
at the bottom of the absorber.
Solution
From Ap. D, b\ trial-and-error:
c)
At the top of the column:
From Ap. D, b\ trial-and-error:
Average mass-transfer coefficients
AYeUage heighW
CalcXlaWe NWG
IniWial eVWimaWe
6.1a. Flash vapori]ation of a heptane-octane mi[ture.
A liquid mixture containing 50 mole % n-heptane (A), 50 mole % n-octane (B), at 303 K, is to be
continuously flash-vaporized at a pressure of 1 atm to vaporize 30 mole % of the feed. What will
be the composition of the vapor and liquid and the temperature in the separator if it behaves as
an ideal stage?
Solution
From the program in Figure 6.3,
6.2a. Flash vapori]ation of a heptane-octane mi[ture.
A liquid mixture containing 50 mole % n-heptane (A), 50 mole % n-octane (B), at 303 K, is to be
continuously flash-vaporized at a temperature of 350 K to vaporize 30 mole % of the feed. What
will be the composition of the vapor and liquid and the pressure in the separator if it behaves as an
ideal stage?
Solution
From the program in Figure 6.3,
6.3d. Flash vapori]ation of a ternar\ mi[ture.
Modify the Mathcad¨ program of Figure 6.3 for ternary mixtures. Test your program with the data
presented in Example 6.2 for a mixture of benzene (A), toluene (B), and o-xylene (C). Critical
temperatures and pressures,and the parameters of the Wagner equation for estimating vapor
pressure (equation 6-5) are included in the following table (Reid, et al., 1987).
Component Tc, K Pc, bar A B C D
benzene 562.2 48.9 ç 6.983 1.332 ç 2.629 ç 3.334
toluene 591.8 41.0 ç 7.286 1.381 ç 2.834 ç 2.792
o-xylene 630.3 37.3 ç 7.534 1.410 ç 3.110 ç 2.860
Solution
Parameters of the Wagner equation, A = benzene, B, toluene, C = o-xylene:
IniWial eVWimaWeV
6.4d. Flash vapori]ation of a ternar\ mi[ture.
CRQVideU Whe WeUQaU\ Pi[WXUe Rf E[aPSOe 6.2 aQd PURbOeP 6.3. EVWiPaWe Whe WePSeUaWXUe, aQd
cRPSRViWiRQ Rf Whe OiTXid aQd YaSRU ShaVeV ZheQ 60% Rf Whe Pi[WXUe haV beeQ YaSRUi]ed aW a cRQVWaQW
SUeVVXUe Rf 1 aWP.
SROXWiRQ
6.5d. Flash vapori]ation of a ternar\ mi[ture.
Consider the ternar\ mi[ture of E[ample 6.2 and Problem 6.3. It is desired to recover in the vapor 75%
of the ben]ene in the feed, and to recover in the liquid 70% of the o-[\lene in the feed. Calculate the
temperature, pressure, fraction of the feed vapori]ed, and the concentration of the 
liquid and gas phases
Solution
6.6b.Batch distillation of a heptane-octane mi[ture.
Repeat the calculations of E[ample 6.3, but for 80 mole % of the liquid distilled.
Solution
Batch distillation of a mi[ture of heptane and octane
IQiWiaO eVWiPaWe
6.8a. Binar\ batch distillation with constant relative volatilit\.
CRQVideU Whe biQaU\ baWch diVWiOOaWiRQ Rf E[aPSOe 6.3. FRU WhiV V\VWeP, Q-heSWaQe ZiWh Q-RcWaQe aW 1
aWP, Whe aYeUage UeOaWiYe YROaWiOiW\ iV D = 2.16 (TUe\baO, 1980). UViQg eTXaWiRQ (6-102) deUiYed iQ PURbOeP
6.7, cRPSXWe Whe cRPSRViWiRQ Rf Whe UeVidXe afWeU 60 PROe% Rf Whe feed iV baWch-diVWiOOed.
SROXWiRQ
IQiWiaO eVWiPaWe:
6.9b. Binar\ batch distillation with constant relative volatilit\.
CRQVideU Whe biQaU\ baWch diVWiOOaWiRQ Rf E[aPSOe 6.3. FRU WhiV V\VWeP, Q-heSWaQe ZiWh Q-RcWaQe aW 1
aWP, Whe aYeUage UeOaWiYe YROaWiOiW\ iV D = 2.16 (TUe\baO, 1980). The Pi[WXUe ZiOO be baWch-diVWiOOed XQWiO
Whe aYeUage cRQceQWUaWiRQ Rf Whe diVWiOOaWe iV 65 PROe % heSWaQe. UViQg eTXaWiRQ (6-102) deUiYed iQ
Problem 6.7, compute the composition of the residue, and the fraction of the feed that is distilled.
Solution
Initial estimates:
6.10b. Mi[tures of light h\drocarbons: m-value correlations.
What is the bubble point of a mi[ture that is 15 mole % isopentane, 30 mole % n-pentane, and 55
mole % n-he[ane? The pressure is 1.0 atm.
Solution
Initial estimate
6.11c. Mi[tures of light h\drocarbons: m-value correlations.
A solution has the folloZing composition, e[pressed as mole percent: ethane, 0.25 %; propane, 25 %;
isobutane, 18.5 %; n-butane, 56 %; isopentane, 0.25 %. In the folloZing, the pressure is 10 bars. Use
equation (6-103) and Table 6.4 to calculate equilibrium distribution coefficients.
a) Calculate the bubble point
Solution
Initial estimate
b) Calculate the deZ 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
Check!
6.12b. Binar\ batch distillation with constant relative volatilit\.
A 30 mole % feed of ben]ene in toluene is to be distilled in a batch operation. A product having an
aYeUage cRPSRViWiRQ Rf 45 PROe % beQ]eQeiV WR be SURdXced. CaOcXOaWe Whe aPRXQW Rf UeVidXe OefW,
aVVXPiQg WhaW D = 2.5 aQd F = 100 PROeV. 
SROXWiRQ
IQiWiaO eVWiPaWeV:
6.13b. Batch distillation of a mi[ture of isopropanol in Zater.
A Pi[WXUe Rf 40 PROe % iVRSURSaQRO iQ ZaWeU iV WR be baWch-diVWiOOed aW 1 aWP XQWiO 70 PROe % Rf Whe
chaUge haV beeQ YaSRUi]ed. CaOcXOaWe Whe cRPSRViWiRQ Rf Whe OiTXid UeVidXe UePaiQiQg iQ Whe VWiOO SRW,
aQd Whe aYeUage cRPSRViWiRQ Rf Whe cROOecWed diVWiOOaWe. VLE daWa fRU WhiV V\VWeP, iQ PROe fUacWiRQ Rf
iVRSURSaQRO, aW 1 aWP aUe (SeadeU aQd HeQOe\, 1998):
T, K 366 357 355.1 354.3 353.6 353.2 353.3 354.5
\ 0.220 0.462 0.524 0.569 0.593 0.682 0.742 0.916
[ 0.012 0.084 0.198 0.350 0.453 0.679 0.769 0.944
CRPSRViWiRQ Rf Whe a]eRWURSe iV [ = \ = 0.685; bRiOiQg SRiQW Rf Whe a]eRWURSe = 353.2 K.
SROXWiRQ
IQiWiaO eVWiPaWe
6.15a. Continuous distillation of a binar\ mi[ture of constant relative volatilit\.
FRU cRQWiQXRXV diVWiOOaWiRQ Rf a biQaU\ Pi[WXUe Rf cRQVWaQW UeOaWiYeYROaWiOiW\, FeQVNe eTXaWiRQ (6-58) caQ
be XVed WR eVWiPaWe Whe PiQiPXP QXPbeU Rf eTXiOibUiXP VWageV UeTXiUed fRU Whe giYeQ VeSaUaWiRQ, NPiQ.
UVe FeQVNe eTXaWiRQ WR eVWiPaWe NPiQ fRU diVWiOOaWiRQ Rf Whe beQ]eQe-WROXeQe Pi[WXUe Rf E[aPSOe 6.4.
AVVXPe WhaW, fRU WhiV V\VWeP aW 1 aWP, Whe UeOaWiYe YROaWiOiW\ iV cRQVWaQW aW D = 2.5.
SROXWiRQ
6.16a. Continuous distillation of a binar\ mi[ture of constant relative volatilit\.
FRU cRQWiQXRXV diVWiOOaWiRQ Rf a biQaU\ Pi[WXUe Rf cRQVWaQW UeOaWiYe YROaWiOiW\, Whe PiQiPXP UefOX[ UaWiR caQ
be deWeUPiQed aQaO\WicaOO\ fURP Whe fROORZiQg eTXaWiRQ (TUe\baO, 1980):
(6-105)
UVe eTXaWiRQ (6-105) WR eVWiPaWe RPiQ fRU diVWiOOaWiRQ Rf Whe beQ]eQe-WROXeQe Pi[WXUe Rf E[aPSOe 6.4.
AVVXPe WhaW, fRU WhiV V\VWeP aW 1 aWP, Whe UeOaWiYe YROaWiOiW\ iV cRQVWaQW aW D = 2.5.
SROXWiRQ
IQiWiaO eVWiPaWe
6.17b. Continuous rectification of a Zater-isopropanol mi[ture.
A ZaWeU-iVRSURSaQRO Pi[WXUe aW iWV bXbbOe SRiQW cRQWaiQiQg 10 PROe % iVRSURSaQRO iV WR be
cRQWiQXRXVO\ UecWified aW aWPRVSheUic SUeVVXUe WR SURdXce a diVWiOOaWe cRQWaiQiQg 67.5 PROe %
iVRSURSaQRO. NiQeW\-eighW SeUceQW Rf Whe iVRSURSaQRO iQ Whe feed PXVW be UecRYeUed. VLE daWa aUe
giYeQ iQ PURbOeP 6.13. If a UefOX[ UaWiR Rf 1.5 WiPeV Whe PiQiPXP iV XVed, hRZ PaQ\ WheRUeWicaO
VWageV ZiOO be UeTXiUed: 
(a) If a SaUWiaO UebRiOeU iV XVed? 
SROXWiRQ
6.24c. A distillation-membrane h\brid for ethanol deh\dration.
MaQ\ iQdXVWUiaOO\ iPSRUWaQW OiTXid V\VWePV aUe difficXOW RU iPSRVVibOe WR VeSaUaWe b\ ViPSOe cRQWiQXRXV
diVWiOOaWiRQ becaXVe Whe ShaVe behaYiRU cRQWaiQV aQ a]eRWURSe, a WaQgeQW SiQch, RU aQ RYeUaOO ORZ UeOaWiYe
YROaWiOiW\. OQe VROXWiRQ iV WR cRPbiQe diVWiOOaWiRQ ZiWh RQe RU PRUe cRPSOePeQWaU\ VeSaUaWiRQ WechQRORgieV
WR fRUP a h\bUid. AQ e[aPSOe Rf VXch a cRPbiQaWiRQ iV Whe deh\dUaWiRQ Rf eWhaQRO XViQg a diVWiOOaWiRQ-
PePbUaQe h\bUid.
IQ a giYeQ aSSOicaWiRQ, 100 PROeV/V Rf a VaWXUaWed OiTXid cRQWaiQiQg 37 PROe % eWhaQRO aQd 63 PROe % ZaWeU
PXVW be VeSaUaWed WR \ieOd a SURdXcW Zhich iV 99 PROe % eWhaQRO, aQd a UeVidXe cRQWaiQiQg 99 PROe %
ZaWeU. The VROXWiRQ ZiOO be fed WR a diVWiOOaWiRQ cROXPQ RSeUaWiQg aW aWPRVSheUic SUeVVXUe, ZiWh a SaUWiaO
UebRiOeU aQd a WRWaO cRQdeQVeU. The UefOX[ UaWiR ZiOO be 1.5 Whe PiQiPXP. The diVWiOOaWe ZiOO eQWeU a PePbUaQe
ZiWh SaUaPeWeUV aP = 70 aQd T = 0.6. The PePbUaQe bRRVWV Whe cRQceQWUaWiRQ VR WhaW Whe SeUPeaWe
VWUeaP iV Whe eWhaQRO-Uich SURdXcW ([P = 0.99). The UeWeQWaWe VWUeaP iV UeWXUQed aV a VaWXUaWed OiTXid WR Whe
cROXPQ WR Whe WUa\ aW Whe QeaUeVW OiTXid cRQceQWUaWiRQ.
a) CaOcXOaWe Whe PROaU fORZ UaWe Rf Whe SURdXcW aQd Rf Whe UeVidXe.
SROXWiRQ
IQiWiaO eVWiPaWeV:
b) CaOcXOaWe Whe PROaU fORZ UaWe aQd cRPSRViWiRQ Rf Whe diVWiOOaWe cRPiQg RXW Rf Whe cROXPQ.
SROXWiRQ
Initial estimates
6.26b. Fenske-UnderZood-Gilliland method.
A distillation column has a feed of 100 kmoles/h. The feed is 10 mole % LNK, 55 mole % LK, and 35
mole % HK and is a saturated liquid. The reflu[ ratio is 1.2 the minimum. We desire 99.5 % recover\ of
the light ke\ in the distillate. The mole fraction of the light ke\ in the distillate should be 0.75. Use the
FUG approach to estimate the number of ideal stages required and the optimal location of the feed-stage.
Equilibrium data:
DLNK = 4.0 DLK = 1.0 DHK = 0.75 
Solution
Kilomoles in distillate:
Kilomoles in bottoms:
Determine minimum reflu[: Saturated liquid, q = 1
Initial estimate
DeWeUmine nXmbeU of ideal VWageV aW R = 1.2 Rmin: Gilliland coUUelaWion
IniWial eVWimaWe
KiUkbUide eTXaWion:
IniWial eVWimaWe:
Feed aW VWage 11
6.27b. Fenske-UnderZood-Gilliland method.
OQe hXQdUed NPROeV/h Rf a WeUQaU\ bXbbOe-SRiQW Pi[WXUe WR be VeSaUaWed b\ diVWiOOaWiRQ haV Whe
fROORZiQg cRPSRViWiRQ:
CRPSRQeQW MROe fUacWiRQ ReOaWiYe YROaWiOiW\
A 0.4 5
B 0.2 3
C 0.4 1
a) FRU a diVWiOOaWe UaWe Rf 60 NPROeV/h, fiYe WheRUeWicaO VWageV, aQd WRWaO UefOX[, caOcXOaWe Whe diVWiOOaWe
aQd bRWWRPV cRPSRViWiRQ b\ Whe FeQVNe eTXaWiRQ.
SROXWiRQ
IQiWiaO eVWiPaWeV:
b) Using the separation in part a) for components A and C, determine the minimum reflu[ ratio b\ the
UnderZood equation.
Solution
Initial estimate:
Initial estimate:
Initial estimate:
c) For a reflu[ ratio of 1.2 times the minimum, determine the number of theoretical stages
required, and the optimum feed location.
Solution
IQiWiaO eVWiPaWe
KiUNbUide eTXaWiRQ:
IQiWiaO eVWiPaWe:
Feed aW VWage 5
6.35c Flash Calculations: the Rachford-Rice Method for Ideal Mi[tures.
a) ShRZ WhaW Whe SURbOeP Rf fOaVh YaSRUi]aWiRQ Rf a PXOWicRPSRQeQW ideaO Pi[WXUe caQ be UefRUPXOaWed aV
VXggeVWed b\ RachfRUd aQd Rice (DRheUW\ aQd MaORQe, CRQceSWXaO DeVigQ Rf DiVWiOOaWiRQ S\VWePV,
McGUaZ-HiOO, NeZ YRUN, NY, 2001):
b) Solve E[ample 6.2 using the Rachford-Rice method.
Solution
Initial estimate:
7.2 Single e[traction.
Appendi[ G-1: Single-Stage E[traction
IniWial eVWimaWeV:
7.3. Single-stage e[traction: insoluble liquids�
7.4. Single-stage e[traction: insoluble liquids�
7.5 Multistage crosscurrent e[traction: insoluble liquids.
ReaUUaQgiQg eTXaWiRQ (3-60)
SRlYeQW flRZ UaWe SeU VWage
TRWal VRlYeQW flRZ UaWe
7.6 Multistage crosscurrent e[traction.
Appendi[ G-2: Multistage Crosscurrent E[traction
Data presented are from E[ample 7-3: acetone-Zater-chloroform s\stem at 298 K
IniWial eVWimaWeV:
IniWial gXeVVeV:
IniWial gXeVVeV:
Composited e[tract
Check material balances
Acetone
Chloroform
Water
	Prob1-1
	Prob1-2
	Prob1-3
	Prob1-4
	Prob1-5&6
	Prob1-7&8
	Prob1-9
	Prob1-10&11
	Prob1-12&13
	Prob14-16
	Prob1-17&18
	Prob1-19-21
	Prob1-22-23
	Prob1-24
	Prob1-25
	Prob1-26-27
	Prob1-28-29
	Prob1-30-31
	Prob2-1-4
	Prob2-7-8
	Prob2-9-12
	Prob2-13-15
	Prob2-16-18
	Prob2-19-21
	Prob2-22-25
	Prob3-1-6
	Prob3-7-13
	Prob3-14-16
	Prob3-17-18
	Prob3-19-27
	Prob4-1-10
	Prob4-11-14
	Prob4-15-18
	Prob4-19-28
	Prob5-1-8
	Prob5-9-11
	Prob5-12
	Prob5-13-16
	Prob5-17-20
	Prob5-21-22
	Prob6
	Prob7-1-6