3 Column diagram total condenser • liquid/vapor streams inside the column flow counter-current in direct contact with each otherreflux drum(accumulator)enriching sectionrefluxL, xRdistillateD, xD• all three external streams (F, D, B) can be liquids (usual case)temperature• for a binary mixture, the compositions xF, xD, xB all refer to the more volatile componentfeedF, xFxR = xDfeed stagestripping sectionyB ≠ xBVLxD ≠ K xB• to keep the liquid flow rate constant, part of the distillate must be returned to the top of the column as refluxboilupV, yBpartial reboiler• the partial reboiler is the last equilibrium stage in the systembottomsB, xB
4 External mass balanceTMB: F = D + B CMB: F xF = D xD + B xB for specified F, xF, xD, xB, there are only 2 unknowns (D, B)distillateD, xDfeedF, xFDefine recovery and splitsB = F - DbottomsB, xB
5 External energy balance assume column is well-insulated, adiabaticEB: F hF + QC + QR= D hD + B hBdistillateD, xDfeedF, xFF, hF are knownD and B are saturated liquidsso hD, hB are also knownunknowns: QC, QR• need another equationbottomsB, xB
6 Balance on condenser1. Mass balance TMB: V1 = D + L0 CMB: y1 = xD = xR (doesn’t help) unknowns: V1, L0 specify external reflux ratio R = L0/D V1 = D + (L0/D)D = (1 + R)D 2. Energy balance V1H1 + QC = (D + L0)hD = V1hD QC = V1(hD – H1) then calculate QR from column energy balancevaporV1, y1refluxL0, xRdistillateD, xDhD > H1QC < 0QR > 0
7 SplitsSometimes used instead of specifying compositions in product streams.What is the fractional recovery (FR) of benzene in the distillate?What is the fractional recovery (FR) of toluene in the bottoms?Most volatile component (MVC) is benzene:xF = 0.46
8 Calculating fractional recoveries B = F – D = = 339
9 Stage-by-stage analysis Lewis-Sorel method Consider the top of the distillation column:vaporV1, y1V1, V2 are saturated vaporsL0, L1 are saturated liquidsrefluxL0, x0distillateD, xDstage 1V2y2L1x1Which streams have compositions related by VLE?V1, L1They are streams leaving the same equilibrium stage.K1(T1,P) = y1/x1Lecture 6 ends hereHow are the compositions of streams V2 and L1 related?Need to perform balances around stage 1.
11 Relationships for stage 2 TMB: L1 + V3 = L2 + V2 CMB: L1x1 + V3y3 = L2x2 + V2y2 EB: L1h1 + V3H3 = L2h2 + V2H2 VLE: K2(T2,P) = y2/x2 can solve for 4 unknowns (L2, x2, V3, y3)V2,y2L1,x1stage 2L2,x2V3,y3and so on… proceed down the column to the reboiler. Very tedious.Simplifying assumption:If li (latent heat of vaporization) is not a strong function of composition, then each mole of vapor condensing on a given stage causes one mole of liquid to vaporize.Constant Molal Overflow (CMO): vapor and liquid flow rates are constant
12 Constant molal overflow TMB: L1 + V3 = L2 + V2CMO: V3 - V2 = L2 - L1 = 0V3 = V2 = VL2 = L1 = LWe can drop all subscripts on L and V in the upper section of the column (above the feed stage).internal reflux ratio: L/V = constant
13 Rectifying column Feed enters at the bottom, as a vapor. No reboiler required.Can give very pure distillate; but bottoms stream will not be very pure.Mass balance around top of column, down to and including stage j:L, xRD, xDreplace rectifying by enriching here and below – parallel with strippingstage jCMB: Vj+1yj+1 = Ljxj + DxDVj+1,yj+1Lj,xjCMO: yj+1 = (L/V) xj + (D/V) xDD = V - Lyj+1 = (L/V) xj + (1 - L/V) xDF, xFB, xBRelates compositions of passing streams.
14 Lewis analysis of rectifying column Assume CMO (Vj = Vj+1 = V; Lj = Lj-1 = L)Need specified xD; xD = y1Stage 1: use VLE to obtain x1x1 = y1/K1(T1,P)4. Use mass balance to obtain y2y2 = (L/V) x1 + (1 - L/V) xD5. Stage 2: use VLE to obtain x2x2 = y2/K2(T2,P)6. Use mass balance to obtain y3y3 = (L/V) x2 + (1 - L/V) xD7. Continue until x = xB
15 Graphical analysis of rectifying column equation of the operating line:y = (L/V) x + (1 - L/V) xDslope = (L/V)always positive (compare to flash drum)y=xVLE•xD= x0(x0,y1)op. lineplotting the operating line:yint = (1 - L/V) xDfind a second point on the operating line:y = x = (L/V) x + (1 - L/V) xD = xDplot xD on y = xyint•recall: xD = xR = x0; the passing stream is y1• the operating line starts at the point (x0,y1)• the operating line gives the compositions of all passing streams (xj,yj+1)
16 McCabe-Thiele analysis: rectifying column Plot VLE line (yi vs. xi)Draw the y = x linePlot xD on y = xPlot yint = xD (1 – L/V)L/V internal reflux ratio, usually not specifiedinstead, the external reflux ratio (R) is specifiedDraw in the operating line6. Step off stages, alternating between VLE and operating line,starting at (x0,y1) located at y = x = xD, until you reach x = xB7. Count the stages.
17 Ex.: MeOH-H2O rectifying column NEVER “step” over the VLE line.Rectifying column with total condenserSpecifications: xD = 0.8, R = 2Find N required to achieve xB = 0.1y=xVLEstage 1(x1,y1)••xD= x0(x0,y1)op. line•(x1,y2)(x2,y2)•stage 21. Plot VLE line2. Draw y=x line•(x2,y3)3. Plot xD on y=x(x3,y3)•stage 34. Plot yint = xD (1 - L/V)•xBLecture 7 ends hereL/V = R/(R+1) = 2/3yint = xD(1 - L/V)= 0.8/3 = 0.26lowest xB possible for this op. line•yint•5. Draw in operating line6. Step off stages from xD to xB7. Count the stagesN = 3
18 Limiting cases: rectification L/V = 0No reflux!Specifications:xD = 0.8, vary R = L/Dy=xVLEL 0R = L/D 0 NO REFLUXL/V 0stage 1(x1,y1)••xD= x0(x0,y1)Column operates like a single equilibrium stage.(Why bother?)0 ≤ R ≤ 0 ≤ L/V ≤ 12. D 0R = L/D TOTAL REFLUXL/V = R/(R+1) 1(L’Hôpital’s Rule)Operating line is y=xL/V = 1Total reflux!Max. distance between VLE and op. lineMax. separation on each equil. stageCorresponds to Nmin, but no distillate!
19 Minimum reflux ratio L/V = 0 Specifications: xD = 0.8, vary R VLE • y=xVLE••xD= x0(x0,y1)The number of stages N required to reach the VLE-op. line intersection point is .Increasing R = L/DDecreasing DDecreasing xB (for fixed N)•This represents xB,min for a particular R.Rmin for this xBIt also represents Rmin for this value of xB.•xB ,min for this RL/V = 10 ≤ L/V ≤ 10 ≤ R ≤
20 Ractual can be specified as a multiple of Rmin Optimum reflux ratioRmintotal costcapital costRoptoperating (energy) costcost/lb∞ stagesmin. heatrequiredexternal reflux ratio, RRule-of-thumb:1.05 ≤ Ropt/Rmin ≤ 1.25Ractual can be specified as a multiple of Rmin
21 Stripping column Feed enters at the top, as a liquid. No reflux required.Can give very pure bottoms; but distillate stream will not be very pure.Mass balance around bottom of column, up to and including stage k:F, xFD, xDVk,ykLk-1,xk-1stage kCMB: Lk-1xk-1 = Vkyk + BxBCMO: yk = (L/V) xk-1 - (B/V) xBL = V + Byk = (L/V) xk-1 + (1 - L/V) xBB, xB
22 Graphical analysis of stripping column equation of the operating line:y = (L/V) x + (1 - L/V) xBslope = L / Valways positivey=xVLEop. line(xN+1,yN+1)PR•plotting the operating line:y = x = (L/V) x + (1 - L/V) xB = xBplot xB on y = xfinding the operating line slope:•xB= xN(xN+1,yN+2)(recall V/B is the boilup ratio)Where is the partial reboiler? Designate this as stage N+1, with xN+1 = xB.Coordinates of the reboiler: (xN+1,yN+1)
23 McCabe-Thiele analysis: stripping column Plot VLE line (yi vs. xi)Draw the y = x linePlot xB on y = xDraw in the operating line5. Step off stages, alternating between VLE and operating line,starting at (xN+1,yN+2) located at y = x = xB, until you reach x = xD6. Count the stages.
24 Ex.: MeOH-H2O stripping column NEVER step over the VLE line.(0.7,1)•op. lineColumn with partial reboilerSpecifications:xB = 0.07,Find N required to achieve xD = 0.55y=xVLEstage 1(xN-2,yN-2)•stage 2xD,max for this boilup ratio•(xN-1,yN-1) ••(xN-2,yN-2)stage 3•xD1. Plot VLE line(xN,yN) •• (xN-1,yN)2. Draw y=x line3. Plot xB = xN+1 on y = x4. Draw op. line•PR• (xN,yN+1)(xN+1,yN+1)5. Step off stages starting at PR•xB= xN+1(xN+1,yN+2)6. Stop when you reach x = xD7. Count the stages.
25 Limiting cases: stripping Specifications:xB = 0.07, vary boilup ratioy=xVLENO BOILUPBehaves as if the column wasn’t even there.(Why bother?)TOTAL BOILUP2. B 0Operating line is y=xTOTAL BOILUP•PRNO BOILUP•xB= xN+1Max. distance between VLE and op. lineMax. separation on each equil. stageCorresponds to Nmin. But no bottoms product!
26 Minimum boilup ratio Specifications: xB = 0.07, vary boilup ratio yD ,max forthis boilup ratio•y=xVLE•The number of stages N required to reach the VLE-op. line intersection point is .Increasing boilup ratioIncreasing xD (for fixed N)Decreasing BTotal boilupThis represents yD,max for a particular boilup ratio.•PRIt also represents the minimum boilup ratio for this value of yD.No boilup•xB= xN+1
27 McCabe-Thiele analysis of complete distillation column Total condenser, partial reboilerSpecifications:xD = 0.8, R = 2xB = 0.07,Find N requiredLocate feed stagey=xbottom op. lineFeed enterson stage 2VLEstage 1•xB•xD•top op. linestage 2••1. Draw y=x line2. Plot xD and xB on y=x3. Draw both op. lines•PR•4. Step off stages startingat either end, using new op. line as you cross their intersectionNEVER step over the VLE line.5. Stop when you reach the other endpoint6. Count stages7. Identify feed stage
28 Feed conditionChanging the feed temperature affects internal flow rates in the columnVLfeedFIf the feed enters as a saturated liquid, the liquid flow rate below the feed stage will increase:VLIf the feed enters as a saturated vapor, the vapor flow rate above the feed stage will increase:If the feed flashes as it enters the feed stage to form a two-phase mixture, 50 % liquid, both the liquid and vapor flow rates will increase:and
30 Different types of feed quality andsaturated liquid feed q = 1saturated vapor feed q = 0feed flashes to form 2-phase < q < 1mixture, q% liquidsubcooled liquid feed q > 1- some vapor condenses on feed platesuperheated vapor q < 0- some liquid vaporizes on feed plate
31 Equation of the feed line rectifying section CMB:stripping section CMB:intersection of top andbottom operating lines:substitute:Wankat gives slope as q/(q-1)andequation of the feed line:
32 Plotting the feed line y = x = zF where does the feed line intersect y=x?y=xVLEsubcooled liq2-phasesat'd liqy = x = zFsat'd vapor•zFWankat gives slope as q/(q-1)superheated vapfeed type q slope, msat'd liquid q=1 m = sat'd vapor q=0 m = 02-phase liq/vap 0<q<1 m < 0subcooled liq q>1 m > 1superheated vap q<0 0<m<1
33 Ex.: Complete MeOH-H2O column Total condenser, partial reboilerSpecifications:xD = 0.9, xB = 0.04, zF = 0.5, R=1Feed is a 2-phase mixture, 50% liq.Find N and NF,opt.y=x162534•xB•xD•zFN = 6 + PR•••••1. Draw y=x lineOperating lines intersect on stage 4. This is NF,opt.2. Plot xD, xB and zF on y=x•3. Draw feed line, slope = -0.54. Draw top op. line, slope = L/V = 0.5Optimum feed stage has intersection point underneath the ‘step’We can independently specify only 2 of the following 3 variables: R, q, V/B (usually: R, q).5. Draw bottom op. line (no calc. required)PR•6. Step off stages starting at either end, using new op. line as you cross the feed line.Using a non-optimal feed location reduces separation.
34 Feed lines in rectifying/stripping columns rectifying columnstripping column•zF••xD•• yD•••top operatingline•••••••zF•PR•bottom operating lineMay be slightly incompatible with previous description of these column endpoints - check•xB•xBtotal condenser, no reboilersat’d vapor feed, liquid bottomsF and B are passing streamspartial reboiler, no condensersat’d liquid feed, vapor distillateF and V are passing streams
35 Design freedom Fixed q. Vary R: Fixed R. Vary q: heat the feed • xB •xD•zFRmin•xB•xD•zFpinch pointdecrease Rqminpinch pointchoice of R dictates required boilup ratio.You cannot “step” over a pinch point – this would require N = . It corresponds to a position in the column where there is no difference in composition between adjacent stages.
36 Another type of pinch point Ethanol-waterxD = 0.82, xB = 0.07zF = 0.5, q = 0.5Find Rminy=x•xB•xD•zFpinch point1. Draw y=x lineVLE2. Plot xD, xB and zF on y=x3. Draw feed line, slope = q/(q-1)4. Draw top op. line to intersect with feed line on VLE line5. Don’t cross the VLE line!6. Redraw top operating line as tangent to VLE.
37 Additional column inputs/outputs Column with two feeds:bottomsB, xBdistillateD, xDColumn with three products:distillateD, xDbottomsB, xBfeedF, zVLVLfeed 2F2, z2, q2side-streamS, xS or ySz2 > z1and/orq2 > q1V´L´V´L´side-streams must be saturated liquid or vaporfeed 1F1, z1, q1VLVLEach intermediate input/output stream changes the mass balance, requiring a new operating line.
38 Multiple feedstreams Total condenser, partial reboiler Specifications: xD = 0.9, xB = 0.07, z1 = 0.4, z2=0.6Some specified q-valuesR = 1. Find N, NF1,opt, NF2,opty=xVLE1PR2534•xB•xD•z2•z1••••1. Draw y=x line2. Plot xD, z1, z2 and xB on y=x•3. Draw both feed lines•z1 = z24. Draw top op. line, slope = L/V5. Calculate slope of middle operating line, L´/V´, and draw middle operating line•Optimum location for feed 1 is stage 5.Optimum location for feed 2 is stage 3.6. Draw bottom operating line (no calc. required)7. Step off stages startingat either end, using new op. line each time you cross an intersection
39 Slope of middle operating line 2-feed mass balances:TMB: F2 + V´ = L´ + DCMB: F2z2 + V´yj+1 = L´xj + DxDfeed 2F2, z2, q2D, xDmiddle operating line equation:y = (L´/V´)x + (DxD - F2z2)/V´obtain slope from:L´ = F2q2 + L = F2q2 + (R)(D)V´ = L´ + D – F2V´L´stage jside-stream feed-stream with –ve flow rate sat’d liq y = x = xSsat’d vapor y = x = ySside-streamS, xS or ySD, xDNot finishedside-stream mass balances:TMB: V´- L´= D + SCMB: V´yj+1 - L´xj = DxD + SxSV´L´stage jmiddle operating line equation:y = (L´/V´)x + (DxD + SxS)/V´
40 McCabe-Thiele analysis of side-streams Saturated liquid side-stream, xs = 0.64Saturated vapor side-stream, ys = 0.73VLEy=x•y=x•xB•xD•xS•z•xB•xD•yS•z••••VLESide-stream must correspond exactly to stage position.
41 Partial condensers y=x • A partial condenser can be used when a vapor distillate is desired:PC12•yD••D, yDVLL, x0V, y1A partial condenser is an equilibrium stage.CMB: Vyj+1 = Lxj + DyDOperating line equation:y = (L/V)x + DyD = (L/V)x + (1 - L/V)yD
42 Total reboilersA total reboiler is simpler (less expensive) than a partial reboiler and is used when the bottoms stream is readily vaporized:y=xTRNN-1•B, xBVLstage NV, yB••xB,yBA total reboiler is not an equilibrium stage.
43 Stage efficiencyUnder real operating conditions, equilibrium is approached but not achieved: Nactual > Nequiloverall column efficiency: Eoverall = Nequil/NactualEfficiency can vary from stage to stage.Reboiler efficiency ≠ tray efficiencyMurphree vapor efficiency:Can also define Murphree liquid efficiency:where yn* is the equilibrium vapor composition (not actually achieved) on stage n:yn* = Kn xnxj* = yj / Kj
44 Ex.: Vapor efficiency of MeOH-H2O column Total condenser, partial reboilerSpecifications:xD = 0.9, xB = 0.07, z = 0.5, q = 0.5,R = 1, EMV,PR = 1, EMV = 0.75.Find N and NF,opt.y=x••xB•xD•z28364517•••••1. Draw y=x line•2. Plot xD, z, and xB on y=x•3. Draw feed line4. Draw top op. line, slope = L/V5. Draw bottom operating line (no calc. required)N = 8 + PRPR•NF,opt = 66. Find partial reboiler7. Step off stages, using EMV to adjust vertical step size.8. Label real stages.To use ELV, adjust horizontal step size instead.
45 Intermediate condensers and reboilers Intermediate condensers/reboilers can improve the energy efficiency of column distillation:by decreasing the heat that must be supplied at the bottom of the column, providing part of the heat using an intermediate reboiler instead- use a smaller (cheaper) heating element at the bottom of the column, or lower temperature steam to heat the boilupby decreasing the cooling that must be supplied at the top of the column, providing part of the cooling using an intermediate condenser instead- use a smaller (cheaper) cooling element at the top of the column, and/or a higher temperature coolant for the intermediate condenserdistillateD, xDbottomsB, xBfeedF, zVLV´L´S, xSintermediate reboileryS = xSV´´L´´VLEach column section has its own operating line.
46 Subcooled refluxIf the condenser is located below the top of the column, the reflux stream has to be pumped to the top of the column.V2L1L0, x0V1, y1D, xDstage 1cPumping a saturated liquid damages the pump, by causing cavitation. The reflux stream (L0) should be subcooled. This will cause some vapor to condense.V1 = V2 - c and L1 = L0 + cCMO is valid below stage 1. Find L/V = L1/V2?EB: V2H2 + L0h0 = V1H1 + L1h1where H1 H2 = H, but h0 ≠ h1 = h(V2 – V1)H = L1h - L0h0cH = (L0 + c)h - L0h0 = L0(h - h0) + chq0 quality of refluxSubcooled reflux causes L/V to increase.Superheated boilup causes L/V to increase.where L0/V1 = (L0/D)/(1 + L0/D) = R/(R + 1)
47 Open steam distillation If the bottoms stream is primarily water, then the boilup is primarily steam.Can replace reboiler with direct steam heating (S).L, xRD, xDmostly MeOHMeOH/H2OfeedF, zTop operating line and feed lines do not change.Bottom operating line is different:stage jVj+1,yj+1Lj,xjTMB: V + B = L + Susually 0CMB: V yj+1 + B xB = L xj + S ySmostly H2ObottomsB, xBCMO: B = LS, ySB, xBOperating line equation:y = (L/V) x - (L/V) xBxint: x = xB
48 Ex.: Open steam distillation of MeOH/H2O Specifications:xD = 0.9, xB = 0.07, zF = 0.5Feed is a 2-phase mixture, 50% liq.Total condenser, open steam, R = 1.Find N and NF,opt.y=xVLE162534•xD•zF•••1. Draw y=x line•2. Plot xD and zF on y=x•3. Plot xB on x-axis4. Draw feed line, slope = q/(q-1)5. Draw top op. line, slope = L/VAll stages are on the column (no partial reboiler).•6. Draw bottom op. line (no calc. required)7. Step off stages startingat either end, using new op. line as you cross their intersectionN = 6 NF,opt = 4•xB
49 Column internals Sieve tray Also called a perforated tray Simple, cheap, easy to cleanGood for feeds that contain suspended solidsPoor turndown performance (low efficiency when operated below designed flow rate); prone to “weeping”
50 Other types of trays Valve tray Bubble cap tray Some valves close when vapor velocity drops, keeping vapor flow rate constantBetter turndown performanceSlightly more expensive, and harder to clean than sieve trayExcellent contact between vapor and liquidRisers around holes prevent weepingGood performance at high and low liquid flow ratesVery expensive, and very hard to cleanNot much used anymore
51 Downcomers Dual-flow tray (no downcomer) • Both liquid and vapor pass through holes• Narrow operating rangeIn large diameter columns, use multi-pass trays to reduce liquid loading in downcomersCross-flow tray (single pass)vertical downcomeralternates sidesDual-pass trayAdd pipe downcomerfour-pass tray
52 Tray efficiencydesign pointexcessive entrainmentefficiencyinefficient mass transferfloodingweeping/dumpingvapor flow rateWeeping/dumping: when vapor flow rate is too low, liquid drips constantly/periodically through holes in sieve trayFlooding: when vapor flow rate is too high, liquid on tray mixes with liquid on tray above
53 Column distillation videos Normal column operation: Flooding: Weeping:
54 Column flooding 1. jet flood (due to entrainment) 3. insufficient downcomer clearance• weir height above downcomer• liquid backs up downcomer2. lack of downcomer seal• weir height below downcomer• vapor flows up downcomer• vapor flow rate too high• ensure bottom edge of downcomer is 1⁄2´´ below top edge of outlet weir.
55 Column sizing 1. Calculate vapor flood velocity, uflood (ft/s) where Csb,f is the capacity factor, from empirical correlation with flow parameter, FPwhere WL and WV are the mass flow rates of liquid and vapor, respectively2. Determine net area required for vapor flow, Anet, based on operating vapor velocity, uop, ft/swhere V is molar vapor flow rate and MWV is average molecular weight of vapor
57 Column sizing, cont.Relationship between net area for vapor flow, Anet, in ft2, and column diameter, D, in ft:where h is the fraction of the cross-sectional area available for vapor flow (i.e., not occupied by the downcomer)The required column diameter, D, in ft, is also:Required column diameter changes where the mass balance changes.- build column in sections, with optimum diameter for each section, or- build column with single diameter:if feed is saturated liquid, design for the bottomif feed is saturated vapor, design for the top - balance section diameters (2-enthalpy feed, intermediate condenser/reboiler)
58 Packed columns structured packing: random packing: • larger surface area, for better contact between liquid and vapor• preferred for column diameters < 2.5´• packing is considerably more expensive than trays• change in vapor/liquid composition is continuous (unlike staged column)• analysis like a staged column: HETP (= Height Equivalent to a Theoretical Plate/Tray)packing height required = no. equil. stages x HETP• packing can be metal, ceramic, glass (depends on feed requirements: corrosive, high T, etc)