Solution Manual For Separation Process Engineering: Includes Mass Transfer Analysis, 3rd Edition

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17SPE 3rdEdition Solution Manual Chapter 1New Problems and new solutions are listed as new immediately after the solution number. These newproblems in chapter 1 are:1A3, 1A4, 1B2-1B4, 1D1.A2.Answers are in the text.A3.New problem for 3rdedition.Answer is d.B1.Everything except some food products has undergone some separation operations. Even thewater in bottles has been purified (either by reverse osmosis or by distillation).B2.New problem for 3rdedition.Many homes have a water softener (ion exchange), or a filter, or acarbon water “filter” (actually adsorption), or a reverse osmosis system.B3.New problem for 3rdedition.For example: the lungs are a gas permeation system, the intestinesand kidney are liquid permeation or dialysis systems.B4.New problem for 3rdedition.You probably used some of the following: chromatography,crystallization, distillation, extraction, filtration and ultrafiltration.D1.New problem for 3rdedition.Basis 1kmol feed..4 kmole E.4MW4618.4 kg10.8 kg.6 kmol Water.6MW18total29.2 kgWeight fraction ethanol = 18.4/29.2 = 0.630Flow rate = (1500 kmol/hr)[(29.2kg)/(1 kmol)] = 43,800 kg/hr.

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18SPE 3rdEdition Solution Manual Chapter 2.New Problems and new solutions are listed as new immediately after the solution number. These newproblems are: 2A6, 2A9 to 2A16, 2C4, 2C8, 2C9, 2D1.g, 2.D4, 2D10, 2D13, 2D24 to 2D30, 2E1, 2F4,2G4 to 2G6, 2H1 to 2H3.2.A1.Feed to flash drum is a liquid at high pressure. At this pressure its enthalpy can be calculatedas a liquid. eg.highLIQF,PpFrefhTcTT. When pressure is dropped the mixture is aboveits bubble point and is a two-phase mixture (It “flashes”).In the flash mixture enthalpy isunchanged but temperature changes.Feed location cannot be found from TFand z on thegraph because equilibrium data is at a lower pressure on the graph used for this calculation.2.A2.Yes.2.A4.2.A6.New Problem.In a flash drum separating a multicomponent mixture, raising the pressure will:i. Decrease the drum diameter and decrease the relative volatilities.Answer is i.2.A8.a. K increases as T increasesb. K decreases as P increasesc. K stays same as mole fraction changes (T, p constant)-Assumption is no concentration effect in DePriester chartsd. K decreases as molecular weight increases2.A9.New Problem.The answer is0.222.A10.New Problem.The answer is b.2.A11.New Problem.The answer is c.1.0.501.0.50xwFlashoperatinglineywEquilibrium(pure water)2.A4zw= 0.965

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192.A12.New Problem.The answer is b.2.A13.New Problem.The answer is c.2.A14.New Problem.The answer is a.2.A15.New Problem.a.The answer is3.5 to 3.6b.The answer is36ºC2.A16.New Problem.The liquid is superheated when the pressure drops, and the energy comes from theamount of superheat.2.B1.Must be sure you don’t violate Gibbs phase rule for intensivevariables in equilibrium.Examples:drumdrumF, z, T, PFF, T , z, pFF, h , z, pdrumF, z, y, PFF, T , z, yFF, h , z, ydrumF, z, x, pFF, T , z, xetc.drumF, z, y, pFdrumdrumF, T , z, T, pdrumF, z, x, TFF, T , y, pDrum dimensions,drumdrumz, F, pFdrumF, T , y, TDrum dimensions,drumz, y, pFF, T , x, petc.FdrumF, T , x, TFF, T , y, x2.B2.This is essentially the same problem (disguised) as problem 2-D1c and e but with an existing(larger) drum and a higher flow rate.With y = 0.58, x = 0.20, and V/F = 0.25 which corresponds to 2-D1c.Iflb moleF1000, D.98 and L2.95 ft from Problem 2-D1ehr.Since D αVand for constant V/F, V α F, we have D αF.With F = 25,000:newoldnewoldnewnewFF= 5, D= 5 D= 4.90, and L= 3 D= 14.7.Existing drum is too small.Feedratedrumcanhandle:FαD2.22existingexistFD41000.98.98givesexistingF16,660 lbmol/hAlternativesa) Do drums in parallel. Add a second drum which can handle remaining 8340 lbmol/h.b)Bypass with liquid mixing

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20Since x is not specified, use bypass. This produces less vapor.c)Look at Eq. (2-62), which becomesvdrumLvvVMWD3K3600Bypass reduces Vc1)Kdrumis already 0.35. Perhaps small improvements can be made with a better demister→ Talk to the manufacturers.c2)ρvcan be increased by increasing pressure. Thus operate at higher pressure. Note thiswill change the equilibrium data and raise temperature.Thus a complete newcalculation needs to be done.d)Try bypass with vapor mixing.e)Other alternatives are possible.2.C2.ABBAzzVFK1K12.C5.a. Start withiiiFzxand let VFLLVKiiiiiiFzzxor xLLLFL K1KFFTheniiiiiiK zyK xLL1KFFFromiiiiiK1 zyx0 we obtain0LL1KFFV = .25 (16660) = 4150LTotalxy = .58,834016,66025,000

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212.C7.iizV1fVF1K1FFrom data in Example 2-2 obtain:V/F0.1.2.3.4.5.6.7.8.91.0f0-.09-.1-.09-.06-.007.07.16.3.49.772.C8.New Problem.drumpxyzdrumTF = L + VzFLxVySolve for L & VOr use lever arm-rule

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222.C9.New Problem.Derivation of Eqs. (2-62) and (2-63). Overall and component mass balances are,12FVLLi1i,L12i,L2iFzL xLxVySubstituting in equilibrium Eqs. (2-60b) and 2-60c)i1i,L1 L2i,L22i,L2iVL2i,L2FzL KxL xVKxSolving,iii,L 21i,L 22i,VL 21i,L1L 21i,VL 2FzFzxL KLVKL KFVLVKDividing numerator and denominator by F and collecting terms.ii,liq 21i,L1L 2i,VL 2zxLV1K1K1FFSinceii,VL2i,L2yKx,i,VL 2ii1i,L1L 2i,VL 2KzyLV1K1K1FFStoichiometric equations,CCCCi,L 2iii,L 2i1i1i1i1x1 ,y1 , thus,yx0which becomesCi,VL 2ii11i,L1L 2i,VL 2K1 z0LV1K1K1FF(2-62)Sincei,L1L 2ii,liq1i,L1L 2i,liq 2i,liq11i,L1L 2i,VL 2KzxKx, we have xLV1K1K1FFIn addition,Ci,L1L 2ii,liq1i,liq 2i11i,L1L 2i,VL 2K1 zxx0LV1K1K1FF(2-63)2.D1.a.V0.4 10040 and LFV60 kmol/hSlope op. lineL V3 2, yxz0.6See graph.y0.77 and x0.48b.V0.41500600 and L900. Rest same as part a.c.Plotx0.2on equil. Diagram andint erceptyxz0.3. yzF V1.2V Fz 1.20.25. From equily0.58.d.Plotx0.45on equilibrium curve.

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23LFV1V F.8Slope4VVV F.2Plot operating line,yxz at z0.51. From mass balanceF37.5kmol/h.e.Find Liquid Density.LmmwwMWxMWxMW.232.04.818.0120.82Then,wmLmwmwMWMW32.0418.01Vxx.2.822.51 ml/mol.79141.00LLLMWV20.82 22.510.925 g/mlFind Vapor Density.vvPMWRT(Need temperature of the drum)vmwmwMWyMWyMW.58 32.04.42 18.0126.15g/molFind Temperature of the Drum T:From Table 3-3 find T corresponding toy.58, x20, T=81.7 C354.7K4vml atm1 atm26.15 g/mol82.0575354.7 K8.9810g/mlmol KFind Permissible velocity:

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24permdrumLvv234drumlvlvlvlvuKKexpABnFCnFDnFEnF2-60SinceVVF0.25 1000250 lbmol/h,FvvlbWV MW25026.156537.5 lb / hlbmolLLLFV1000250750 lbmol/h, and WLMW75020.8215, 615 lb/h,4VLlvlvVLW156158.8910F0.0744, andnF2.598W6537.5.925Then4drumperm4.9258.9810K.442, and u.44214.19 ft/s8.9810v2cs43permvVMW250 26.15454 g/lbA2.28 ft .u360014.1936008.9810g/ml28316.85 ml/ftThus,csD4A1.705ft. Use 2 ft diameter.L ranges from3D6 ft to 5D=10 ftNote that this design is conservative if a demister is used.f.Plot T vs x from Table 3-3.WhenT77 C, x0.34, y0.69.This problem is nowvery similar to 3-D1c.Can calculate V/F from mass balance,FzLxVy.This isVzy0.40.34FzFV xVy or0.17Fyx0.690.34g.Part g is a new problem.V = 16.18 mol/h, L = 33.82, y= 0.892, x = 0.756.2-D2.Work backwards. Starting with x2, find y2= 0.62 from equilibrium. From equilibrium pointplotop.lineofslope222VL V1V F3 7.FThisgives21z0.51x(see Figure). Then from equilibrium,1y0.78.For stage 1,1111zxV0.550.510.148Fyx0.780.51.2.D3.a.z0.4V F0.6V6.0 k mol h,L4.0Op. eq.LyxzV FV2yx2 33See graph:My0.55x0.18T ~ 82.8 Clinear interpretation on Table 2-7.

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25b.Product78.0 Cx0.30,y0.665,Mass Bal:FzLxVyFV xVyor4.010V 0.30.665VV2.985 and V F0.2985Can also calculate V/F from slope.c.VF10,0.3V3 & L7FLz7zyxxVV Fz0.3Ify0.8,x0.545@ equilThen7z0.30.80.5450.6215.3Can also draw line of slope73through equil point.

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262.D4.New problem in 3rdedition.Highest temperature is dew pointV F0Setiiiiizy .KyxWantiiixyK1.0refNewrefOldiiKTKTyKIf pick C4 as reference: First guessbu tan eK1.0,T41 C:C3C6K3.1,K0.125iiy.2.35.454.0145K3.11.0.125T too lowGuess for referenceC4K4.014T118 C:C3C6K8.8,K.9iiy.2.35.450.6099K8.84.0145.9C4,NEWK4.0145 .60992.45,T85:C2C6K6.0,K0.44iiy.2.35.451.20K62.45.44C4,NEWK2.451.22.94, T96 C :C3C6K6.9,K0.56iiy.2.35.450.804Gives 84 CK6.92.94.56

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27Use 90.5º → Avg last two TC4C3C6K2.7,K6.5,K0.49ii.2.35.45yK1.0796.52.7.49T ~ 8788º CNote: hexane probably better choice as reference.2.D5.a)b)1111LFyxzVVPlot 1stOp line.y = x = z = 0.55to x1= 0.3 on eq. curve (see graph)SlopeL0.550.80.250.454545V.550.5511111L0.45454 & LVF1000Vc)Stage 22VL0.75F0.25 ,3 ,yxz0.66.FV0.25FPlot op lineAt20.66Fz0.66x0, yzV F2.64. At y0,xz0.880.25LL F0.75From graph22y0.82, x0.63.222VVF0.25 687.5F171.875 kmol/hF1= 1000z1= 0.55p1,2= 1 atmx2x1= 0.30v2y2v1= F2y1= z2212V0.25Fy1= 0.66 = z2V1= 687.5 kmol/h = F21V687.50.6875F1000

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282.D6.RR eq.,iiiK1 z01K1 V F, First guess V/F = 0.611.4 .450.2.350.7.2f0.021511.4 .610.2 0.610.7 0.6Use Newtonian Convergence2ciik2i1iK1zdfdV F1K1 V Fkk1kfVVdfFFd V FT = 50ºCP = 200 kPaKc4= 2.4zc4= 0.45F = 1.0 kmol/minzc5= 0.35Zc6= 0.20LVKc5= 0.80Kc6= 0.30

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2922212221.40.450.20.350.70.20df0.570V11.4 .610.2 0.610.7 0.6dF2V02150.60.6377F0.57021.4 .450.2 .350.70.2f0.0002811.4 0.637710.2 0.637710.7 0.6377Which is close enough.iiiic 4c 4iyK xz0.45x0.2377,Vy2.40.23770.570511.4 .63771K1Fc5c50.35x0.4012, y0.8 0.40120.321010.2 0.6377c68ii0.200.36130.1084x, y0.30 0.401210.7 0.6377x1.0002y0.99982.D7.ABBAzzVFK1K1MPK5.6 and K0.21V0.30.70.2276F0.2115.61Eq. (2-38)MMMz0.3x0.1466V14.60.22761K1FPMx1x0.8534,MMMyKx5.60.14660.8208PMy1y0.17922.D8.Use Rachford-Rice eqn:iiiK1 zVf0F1K1 V / F.Note that 2 atm = 203 kPa.FindiKfrom DePriester Chart:123K73,K4.1 K.115Converge onV F.076,VF V F152 kmol/h, LFV1848 kmol/h.FromiiizxV1K1Fwe obtain123x.0077,x.0809,x.9113Fromiii123yK x , we obtain y.5621,y.3649,y.10482.D9.Need hFto plot on diagram. Since pressure is high, feed remains a liquidLFPFrefrefhCTT,T0from chartLEtOHwPPEtOHPwCCxCx

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30WhereEtOHwxand xaremolefractions. Convert weight to mole fractions.Basis:100 kg mixture3030 kg EtOH0.651 kmol46.0770 kg water70 18.0163.885Total = 4.536 kmolAvg.100MW22.0464.536Mole fracs:Ew0.6512x0.1435, x0.85654.536.UseLEtOHPPCat 100 C as an average Cvalue.LPkcalC37.96 .143518.0 .856520.86 kmol CPer kg this isLPavgC20.86kcal0.946MW22.046kgCFh0.946 2000189.2 kcal/kgwhich can now be plotted on the enthalpy composition diagram.ObtaindrumEET88.2 C,x0.146, and y0.617.ForF1000find L and V from F = L + V andFzLxVywhich gives V = 326.9, and L = 673.1Note: If use wt. fracs.LLPPavgC23.99 & CMW1.088Fand h217.6.Allwrong.

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312.D.10New Problem. Solution400 kPa, 70ºCC4z35Mole % n-butaneC6x0.7From DePriester chartC3C4C6K5,K1.9,K0.3KnowiiiiiiiiizyK x ,x,xy1zV1K1FR.R.iiC3C6C 4C6iK1 z0z1zz.65zV1K1FFor C6C6C6C6C6zzV0.7z0.7 10.7VVF1K110.7FFC6Vz0.70.49 FRR Eq:C6C64 .65z0.9 .350.7z0VVV1410.910.7FFF2 equations & 2 unknowns. Substitute in forC6z. Do in Spreadsheet.Use Goal–Seek to findV F.V0.594Fwhen R.R. equation0.000881.C6Vz0.70.490.7(0.49)(0.594)0.40894F2.D11.L F0.6V F0.4 & L V1.5Operating line:Slope1.5, through yxz0.4

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