Report

Continuous Column Distillation Column diagram total condenser • to keep the liquid flow rate constant, part of the distillate must be returned to the top of the column as reflux • the partial reboiler is the last equilibrium stage in the system feed F, xF feed stage stripping section • for a binary mixture, the compositions xF, xD, xB all refer to the more volatile component reflux L, xR L V temperature • all three external streams (F, D, B) can be liquids (usual case) reflux drum (accumulator) enriching section • liquid/vapor streams inside the column flow countercurrent in direct contact with each other distillate D, xD xR = xD yB ≠ xB xD ≠ K xB boilup V, yB partial reboiler bottoms B, xB External mass balance TMB: CMB: F=D+B F x F = D xD + B xB for specified F, xF, xD, xB, there are only 2 unknowns (D, B) distillate D, xD feed F, xF æx -x ö B D = çç F ÷÷ F è xD - xB ø B=F-D bottoms B, xB External energy balance • assume column is wellinsulated, adiabatic EB: distillate D, xD F hF + QC + QR = D hD + B hB F, hF are known feed F, xF D and B are saturated liquids so hD, hB are also known unknowns: QC, QR • need another equation bottoms B, xB Balance on condenser 1. Mass balance TMB: V1 = D + L 0 CMB: y1 = xD = xR (doesn’t help) unknowns: V1, L0 specify external reflux ratio R = L0/D vapor V1, y1 reflux L 0 , xR distillate D, xD 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 balance hD > H1 QC < 0 QR > 0 Splits Sometimes 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 FRMVC = xDD 0.99D = xF F 0.46F FRLVC = (1- xB )B 0.98D = (1- xF )F 0.54F Calculating fractional recoveries æx -x ö æ 0.46 - 0.02 ö 0.44 F B D = çç ÷÷ F = ç F = (620) = 281 ÷ 0.97 è 0.99 - 0.02 ø è xD - xB ø B = F – D = 620 - 281 = 339 FRMVC FRLVC xDD 0.99(281) = = = 0.975 xF F 0.46(620) (1- xB )B 0.98(339) = = = 0.992 (1- xF )F 0.54(620) Stage-by-stage analysis Lewis-Sorel method Consider the top of the distillation column: vapor V1, y1 stage 1 L1 x1 V2 y2 V1, V2 are saturated vapors L0, L1 are saturated liquids reflux L0, x0 distillate D, xD Which streams have compositions related by VLE? V1, L1 They are streams leaving the same equilibrium stage. K1(T1,P) = y1/x1 How are the compositions of streams V2 and L1 related? Need to perform balances around stage 1. Relationships for stage 1 vapor V1, y1 stage 1 L1 x1 reflux L0, x0 distillate D, xD V2 y2 There are 14 variables: 4 flow rates (L1, V2, L0, V1) 4 compositions (x1, y2, x0, y1) 4 enthalpies (h1, H2, h0, H1) T1, P TMB: CMB: EB: VLE: L0 + V2 = L1 + V1 L0x0 + V2y2 = L1x1 + V1y1 L0h0 + V2H2 = L1h1 + V1H1 K1(T1,P) = y1/x1 We usually specify 10 of them: P, xD, D, R = L0/D xD = x0 = y1 V1 = L0 + D T1 and all 4 enthalpies (by VLE) 4 unknowns (L1, x1, V2, y2) and 4 equations: problem is completely specified. Relationships for stage 2 L1,x1 V2,y2 stage 2 L2,x2 V3,y3 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) and 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 Constant molal overflow TMB: L1 + V3 = L2 + V2 CMO: V3 - V2 = L2 - L1 = 0 V3 = V2 = V L2 = L1 = L We 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 Rectifying column Feed enters at the bottom, as a vapor. No reboiler required. L, xR D, xD 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: stage j Vj+1,yj+1 F, xF CMB: Vj+1yj+1 = Ljxj + DxD Lj,xj B, xB CMO: yj+1 = (L/V) xj + (D/V) xD D=V-L yj+1 = (L/V) xj + (1 - L/V) xD Relates compositions of passing streams. Lewis analysis of rectifying column 1. Assume CMO (Vj = Vj+1 = V; Lj = Lj-1 = L) 2. Need specified xD; xD = y1 3. Stage 1: use VLE to obtain x1 x1 = y1/K1(T1,P) 4. Use mass balance to obtain y2 y2 = (L/V) x1 + (1 - L/V) xD 5. Stage 2: use VLE to obtain x2 x2 = y2/K2(T2,P) 6. Use mass balance to obtain y3 y3 = (L/V) x2 + (1 - L/V) xD 7. Continue until x = xB Graphical analysis of rectifying column equation of the operating line: y = (L/V) x + (1 - L/V) xD slope = (L/V) always positive (compare to flash drum) •xD = x0 (x0,y1) plotting the operating line: yint = (1 - L/V) xD find a second point on the operating line: y = x = (L/V) x + (1 - L/V) xD = xD plot xD on y = x yint• 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) McCabe-Thiele analysis: rectifying column 1. 2. 3. 4. Plot VLE line (yi vs. xi) Draw the y = x line Plot xD on y = x Plot yint = xD (1 – L/V) L/V internal reflux ratio, usually not specified instead, the external reflux ratio (R) is specified L L L D = R = D= (L + D) V V R +1 D D 5. Draw in the operating line 6. Step off stages, alternating between VLE and operating line, starting at (x0,y1) located at y = x = xD, until you reach x = xB 7. Count the stages. Ex.: MeOH-H2O rectifying column NEVER “step” over the VLE line. Rectifying column with total condenser Specifications: xD = 0.8, R = 2 Find N required to achieve xB = 0.1 stage 1 (x1,y1)• stage 2 (x2,y2)• 1. Plot VLE line 2. Draw y=x line 3. Plot xD on y=x 4. Plot yint = xD (1 - L/V) L/V = R/(R+1) = 2/3 yint = xD(1 - L/V)= 0.8/3 = 0.26 stage 3 (x3,y3)• yint• •xD= x0 (x0,y1) •(x1,y2) •(x2,y3) • 5. Draw in operating line lowest xB possible for this op. line 6. Step off stages from xD to xB 7. Count the stages N=3 •xB Limiting cases: rectification Specifications: xD = 0.8, vary R = L/D 1. L 0 R = L/D 0 NO REFLUX L/V 0 Column operates like a single equilibrium stage. (Why bother?) L/V = 0 No reflux! stage 1 (x1,y1)• 0≤R≤ 0 ≤ L/V ≤ 1 2. D 0 R = L/D TOTAL REFLUX L/V = R/(R+1) 1 (L’Hôpital’s Rule) Operating line is y=x Max. distance between VLE and op. line Max. separation on each equil. stage Corresponds to Nmin, but no distillate! L/V = 1 Total reflux! •xD= x0 (x0,y1) Minimum reflux ratio L/V = 0 Specifications: xD = 0.8, vary R • The number of stages N required to reach the VLE-op. line intersection point is . • • This represents xB,min for a particular R. It also represents Rmin for this value of xB. • •xD= x0 (x0,y1) Increasing R = L/D Decreasing D Decreasing xB (for fixed N) Rmin for this xB 0 ≤ L/V ≤ 1 • xB ,min for this R L/V = 1 0≤R≤ Optimum reflux ratio capital cost operating (energy) cost ∞ stages cost/lb total cost min. heat required Rmin Ropt external reflux ratio, R Rule-of-thumb: 1.05 ≤ Ropt/Rmin ≤ 1.25 Ractual can be specified as a multiple of Rmin Stripping column Feed enters at the top, as a liquid. F, xF No reflux required. D, xD Lk-1, xk-1 Vk,yk stage k 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: CMB: Lk-1xk-1 = Vkyk + BxB B, xB CMO: yk = (L/V) xk-1 - (B/V) xB L=V+B yk = (L/V) xk-1 + (1 - L/V) xB Graphical analysis of stripping column equation of the operating line: y = (L/V) x + (1 - L/V) xB slope = L / V always positive plotting the operating line: y = x = (L/V) x + (1 - L/V) xB = xB plot xB on y = x (xN+1,yN+1) PR • finding the operating line slope: L V +B B = = 1+ V V V •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) McCabe-Thiele analysis: stripping column 1. Plot VLE line (yi vs. xi) 2. Draw the y = x line 3. Plot xB on y = x 4. Draw in the operating line 5. Step off stages, alternating between VLE and operating line, starting at (xN+1,yN+2) located at y = x = xB, until you reach x = xD 6. Count the stages. Ex.: MeOH-H2O stripping column NEVER step over the VLE line. Column with partial reboiler Specifications: xB = 0.07, V / B = 2 Find N required to achieve xD = 0.55 1. Plot VLE line 2. Draw y=x line (0.7,1) • stage 1 (xN-2,yN-2) stage 2 •• (xN-1,yN-1) • •(xN-2,yN-2) stage 3 (xN,yN) • • (xN-1,yN) xD,max for this boilup ratio 3. Plot xB = xN+1 on y = x 4. Draw op. line L / V = 1+ B / V = 1.5 y = 1 = 1.5x - 0.05 x = 1.05 /1.5 = 0.7 PR• (xN+1, yN+1) • (xN,yN+1) 5. Step off stages starting at PR 6. Stop when you reach x = xD 7. Count the stages. • xB= xN+1 (xN+1,yN+2) x •D Limiting cases: stripping Specifications: xB = 0.07, vary boilup ratio V / B 1. NO BOILUP Behaves as if the column wasn’t even there. (Why bother?) 1£ L /V £ ¥ ¥ ³V / B ³ 0 2. B 0 TOTAL BOILUP PR• L /V = 1 TOTAL BOILUP Operating line is y=x L /V = ¥ • xB= xN+1 Max. distance between VLE and op. line Max. separation on each equil. stage Corresponds to Nmin. But no bottoms product! NO BOILUP Minimum boilup ratio Specifications: xB = 0.07, vary boilup ratio yD ,max for this boilup ratio • • • The number of stages N required to reach the VLE-op. line intersection point is . This represents yD,max for a particular boilup ratio. It also represents the minimum boilup ratio for this value of yD. • L /V = 1 Total boilup PR• L /V = ¥ • xB= xN+1 No boilup 1£ L /V £ ¥ ¥ ³V / B ³ 0 McCabe-Thiele analysis of complete distillation column Total condenser, partial reboiler Specifications: xD = 0.8, R = 2 xB = 0.07, V / B = 2 Find N required Locate feed stage Feed enters on stage 2 stage 2 • stage 1 • •xD • 1. Draw y=x line 2. Plot xD and xB on y=x 3. Draw both op. lines 4. Step off stages starting at either end, using new op. line as you cross their intersection 5. Stop when you reach the other endpoint 6. Count stages 7. Identify feed stage PR• • xB • NEVER step over the VLE line. Feed condition • Changing the feed temperature affects internal flow rates in the column • If the feed enters as a saturated liquid, the liquid L V F + L +V = L +V • If the feed enters as a saturated vapor, the vapor flow rate above the feed stage will increase: V =V + F • 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: L = L + 0.5F and V feed F flow rate below the feed stage will increase: L = L+F L V = V + 0.5F Feed quality, q EB: FhF + LhL +VHV = LhL +VHV rearrange: FhF + (V -V)HV = (L - L)hL TMB: V -V = L - L - F substitute: FhF + (L - L - F)HV = (L - L)hL combine terms: (L - L)(HV - hL ) = F(HV - hF ) define: L - L HV - hF º q q mol sat’d liquid = generated on feed F HV - hL plate, per mol feed Different types of feed quality L -L qº F L = L + qF and V = V + (1- q)F saturated liquid feed L = L+F q=1 saturated vapor feed V =V + F q=0 feed flashes to form 2-phase mixture, q% liquid L = L + qF 0<q<1 subcooled liquid feed q>1 - some vapor condenses on feed plate superheated vapor - some liquid vaporizes on feed plate q<0 Equation of the feed line rectifying section CMB: Vy j+1 = Lx j + DxD stripping section CMB: Vy k = Lxk-1 - BxB intersection of top and bottom operating lines: (V -V)y = (L - L)x - (BxB + DxD ) substitute: BxB + DxD = FzF æ L -L ö FzF y = -ç ÷x + èV -V ø V -V L - L = qF and equation of the feed line: zF q y =x+ (1- q) 1- q V -V = (1- q)F Plotting the feed line sat'd liq zF q y =x+ (1- q) 1- q sat'd vapor •zF where does the feed line intersect y=x? zF q x =x+ (1- q) 1- q æ q ö zF x ç1+ ÷= è (1- q) ø 1- q zF x = y = x = zF 1- q 1- q feed type sat'd liquid sat'd vapor 2-phase liq/vap subcooled liq superheated vap q slope, m q=1 m= q=0 m=0 0<q<1 m < 0 q>1 m>1 q<0 0<m<1 Ex.: Complete MeOH-H2O column Total condenser, partial reboiler Specifications: xD = 0.9, xB = 0.04, zF = 0.5, R=1 Feed is a 2-phase mixture, 50% liq. Find N and NF,opt. N = 6 + PR 4 • 3 • 5• 1. Draw y=x line •zF 2. Plot xD, xB and zF on y=x 6 • 3. Draw feed line, slope = -0.5 2• 1 • •xD Operating lines intersect on stage 4. This is NF,opt. 4. Draw top op. line, slope = L/V = 0.5 5. Draw bottom op. line (no calc. required) 6. Step off stages starting at either end, using new op. line as you cross the feed line. PR • • xB We can independently specify only 2 of the following 3 variables: R, q, V/B (usually: R, q). Using a non-optimal feed location reduces separation. Feed lines in rectifying/stripping columns stripping column rectifying column • • • •xD • • • • yD •z F • •z •F •xB • • • bottom operating line total condenser, no reboiler sat’d vapor feed, liquid bottoms F and B are passing streams PR• • top operating line • xB partial reboiler, no condenser sat’d liquid feed, vapor distillate F and V are passing streams Design freedom Fixed q. Vary R: Fixed R. Vary q: Rmin •xD •xD pinch point decrease R •zF • xB choice of R dictates required boilup ratio. •zF qmin pinch point • xB 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. Another type of pinch point Ethanol-water xD = 0.82, xB = 0.07 zF = 0.5, q = 0.5 Find Rmin pinch point 1. Draw y=x line 2. Plot xD, xB and zF on y=x •zF 3. Draw feed line, slope = q/(q-1) 4. Draw top op. line to intersect with feed line on VLE line 5. Don’t cross the VLE line! 6. Redraw top operating line as tangent to VLE. • xB •xD Additional column inputs/outputs Column with two feeds: Column with three products: L V L´ V´ feed 2 F2, z2, q2 feed 1 F1, z1, q1 L distillate D, xD z2 > z1 and/or q2 > q 1 V L V L´ V´ feed F, z L bottoms B, xB V distillate D, xD side-stream S, xS or yS side-streams must be saturated liquid or vapor bottoms B, xB Each intermediate input/output stream changes the mass balance, requiring a new operating line. Multiple feedstreams Total condenser, partial reboiler Specifications: xD = 0.9, xB = 0.07, z1 = 0.4, z2=0.6 Some specified q-values R = 1. Find N, NF1,opt, NF2,opt 3• 2. Plot xD, z1, z2 and xB on y=x •z2 5• •z1 = z2 3. Draw both feed lines •z1 4. Draw top op. line, slope = L/V 7. Step off stages starting at either end, using new op. line each time you cross an intersection •xD 4 • 1. Draw y=x line 5. Calculate slope of middle operating line, L´/V´, and draw middle operating line 6. Draw bottom operating line (no calc. required) 2• 1 • PR • x•B Optimum location for feed 1 is stage 5. Optimum location for feed 2 is stage 3. Slope of middle operating line 2-feed mass balances: TMB: F2 + V´ = L´ + D CMB: F2z2 + V´yj+1 = L´xj + DxD middle operating line equation: y = (L´/V´)x + (DxD - F2z2)/V´ obtain slope from: L´ = F2q2 + L = F2q2 + (R)(D) V´ = L´ + D – F2 feed 2 F2, z2, q2 D, xD stage j L´ V´ side-stream feed-stream with –ve flow rate sat’d liq y = x = xS sat’d vapor y = x = yS D, xD side-stream mass balances: TMB: V´- L´= D + S CMB: V´yj+1 - L´xj = DxD + SxS stage j middle operating line equation: y = (L´/V´)x + (DxD + SxS)/V´ L´ side-stream S, xS or yS V´ McCabe-Thiele analysis of side-streams Saturated liquid side-stream, xs = 0.64 • • Saturated vapor side-stream, ys = 0.73 •xD • • • •yS •xS •z •x B •z •x B Side-stream must correspond exactly to stage position. •xD Partial condensers A partial condenser can be used when a vapor distillate is desired: 2 • D, yD V, y1 L, x0 L V A partial condenser is an equilibrium stage. CMB: Vyj+1 = Lxj + DyD Operating line equation: y = (L/V)x + DyD = (L/V)x + (1 - L/V)yD 1 • PC • •yD Total reboilers A total reboiler is simpler (less expensive) than a partial reboiler and is used when the bottoms stream is readily vaporized: N-1 • V L stage N N• V, yB B, xB A total reboiler is not an equilibrium stage. TR •xB,yB Stage efficiency Under real operating conditions, equilibrium is approached but not achieved: Nactual > Nequil overall column efficiency: Eoverall = Nequil/Nactual Efficiency can vary from stage to stage. Reboiler efficiency ≠ tray efficiency Murphree vapor efficiency: EMV = y n - y n+1 y n * -y n+1 where yn* is the equilibrium vapor composition (not actually achieved) on stage n: yn* = Kn xn Can also define Murphree liquid efficiency: EML = x n - x n-1 x n * -x n-1 xj* = yj / Kj Ex.: Vapor efficiency of MeOH-H2O column Total condenser, partial reboiler Specifications: 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. • 5 • 6 • 1. Draw y=x line • 4 • 3 • 2 • 1 •xD 7 2. Plot xD, z, and xB on y=x • 3. Draw feed line •z 8 4. Draw top op. line, slope = L/V 5. Draw bottom operating line (no calc. required) PR• NF,opt = 6 6. Find partial reboiler 7. Step off stages, using EMV to adjust vertical step size. N = 8 + PR • xB 8. Label real stages. To use ELV, adjust horizontal step size instead. Intermediate condensers and reboilers Intermediate condensers/reboilers can improve the energy efficiency of column distillation: 1. 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 boilup 1. by decreasing the cooling that must be supplied at the top of the column, providing part of the cooling using an intermediate condenser instead 2. - use a smaller (cheaper) cooling element at the top of the column, and/or a higher temperature coolant for the intermediate condenser Each column section has its own operating line. L V L´ V´ distillate D, xD feed F, z S, xS L´´ intermediate reboiler V´´ yS = xS L V bottoms B, xB Subcooled reflux If the condenser is located below the top of the column, the reflux stream has to be pumped to the top of the column. Pumping 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 stage 1 c V2 L0, x0 L1 L1 = L0 + c D, xD CMO is valid below stage 1. Find L/V = L1/V2? c= V1, y1 h - h0 L = (1- q0 )L0 q quality of reflux 0 H -h 0 ( ) 2 - q0 L0 / V1 L1 L0 + c L0 + (1- q0 )L0 = = = V2 V1 + c V1 + (1- q0 )L0 1+ (1- q0 )L0 / V1 where L0/V1 = (L0/D)/(1 + L0/D) = R/(R + 1) EB: V2H2 + L0h0 = V1H1 + L1h1 where H1 H2 = H, but h0 ≠ h1 = h (V2 – V1)H = L1h - L0h0 cH = (L0 + c)h - L0h0 = L0(h - h0) + ch Subcooled reflux causes L/V to increase. Superheated boilup causes L/V to increase. 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, xR D, xD mostly MeOH MeOH/H2O feed F, z Top operating line and feed lines do not change. Bottom operating line is different: stage j TMB: Vj+1, yj+1 S, yS Lj, xj B, xB V + B = L + S usually 0 mostly H2O bottoms B, xB CMB: V yj+1 + B xB = L xj + S yS CMO: B=L Operating line equation: y = (L/V) x - (L/V) xB xint: x = xB Ex.: Open steam distillation of MeOH/H2O Specifications: xD = 0.9, xB = 0.07, zF = 0.5 Feed is a 2-phase mixture, 50% liq. Total condenser, open steam, R = 1. Find N and NF,opt. 3 • 2 • 1 • •xD 4• 1. Draw y=x line 5• 2. Plot xD and zF on y=x •zF 3. Plot xB on x-axis 4. Draw feed line, slope = q/(q-1) 5. Draw top op. line, slope = L/V 6. Draw bottom op. line (no calc. required) 7. Step off stages starting at either end, using new op. line as you cross their intersection 6 • All stages are on the column (no partial reboiler). N=6 x•B NF,opt = 4 Column internals Sieve tray • • • • Also called a perforated tray Simple, cheap, easy to clean Good for feeds that contain suspended solids Poor turndown performance (low efficiency when operated below designed flow rate); prone to “weeping” Other types of trays Valve tray • Some valves close when vapor velocity drops, keeping vapor flow rate constant • Better turndown performance • Slightly more expensive, and harder to clean than sieve tray Bubble cap tray • Excellent contact between vapor and liquid • Risers around holes prevent weeping • Good performance at high and low liquid flow rates • Very expensive, and very hard to clean • Not much used anymore Downcomers Dual-flow tray (no downcomer) • Both liquid and vapor pass through holes • Narrow operating range In large diameter columns, use multi-pass trays to reduce liquid loading in downcomers Cross-flow tray (single pass) Dual-pass tray Tray efficiency efficiency design point flooding weeping/du mping vapor flow rate • Weeping/dumping: when vapor flow rate is too low, liquid drips constantly/periodically through holes in sieve tray • Flooding: when vapor flow rate is too high, liquid on tray mixes with liquid on tray above Column distillation videos Normal column operation: http://www.youtube.com/watch?v=QQgtcNzW9Nw&NR=1 Flooding: http://www.youtube.com/watch?v=tHOlFleAkNE Weeping: http://www.youtube.com/watch?v=tRRxBCSuz48 Column flooding 1. jet flood (due to entrainment) 3. insufficient downcomer clearance 2. lack of downcomer seal • vapor flow rate too high • weir height below downcomer • vapor flows up downcomer • weir height above downcomer • liquid backs up downcomer • ensure bottom edge of downcomer is 1⁄2´´ below top edge of outlet weir. Column sizing 1. Calculate vapor flood velocity, uflood (ft/s) æs ö = Csb,f ç ÷ è 20 ø 0.2 uflood rL - rV rV where Csb,f is the capacity factor, from empirical correlation with flow parameter, FP WL FP = WV rV rL where WL and WV are the mass flow rates of liquid and vapor, respectively 2. Determine net area required for vapor flow, Anet, based on operating vapor velocity, uop, ft/s V (MWV ) uop = 0.75uflood = 3600rV Anet where V is molar vapor flow rate and MWV is average molecular weight of vapor Tray spacing Column sizing, cont. Relationship between net area for vapor flow, Anet, in ft2, and column diameter, D, in ft: p D 2h Anet = 4 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: D= ( 4V MWV ) 3600phrV uop 4V RT » 3600phuop P 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 bottom if feed is saturated vapor, design for the top - balance section diameters (2-enthalpy feed, intermediate condenser/reboiler) 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)