# Continuous Column Distillation

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Continuous Column Distillation

Column diagram total condenser
• liquid/vapor streams inside the column flow counter-current in direct contact with each other reflux drum (accumulator) enriching section reflux L, xR distillate D, 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 component feed F, xF xR = xD feed stage stripping section yB ≠ xB V L xD ≠ K xB • to keep the liquid flow rate constant, part of the distillate must be returned to the top of the column as reflux boilup V, yB partial reboiler • the partial reboiler is the last equilibrium stage in the system bottoms B, xB

External mass balance TMB: F = D + B CMB: F xF = D xD + B xB for specified F, xF, xD, xB, there are only 2 unknowns (D, B) distillate D, xD feed F, xF Define recovery and splits B = F - D bottoms B, xB

External energy balance
assume column is well-insulated, adiabatic EB: F hF + QC + QR = D hD + B hB distillate D, xD feed F, xF F, hF are known 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 + 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 balance vapor V1, y1 reflux L0, xR distillate D, xD 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

Calculating fractional recoveries
B = F – D = = 339

Stage-by-stage analysis Lewis-Sorel method
Consider the top of the distillation column: vapor V1, y1 V1, V2 are saturated vapors L0, L1 are saturated liquids reflux L0, x0 distillate D, xD stage 1 V2 y2 L1 x1 Which streams have compositions related by VLE? V1, L1 They are streams leaving the same equilibrium stage. K1(T1,P) = y1/x1 Lecture 6 ends here How are the compositions of streams V2 and L1 related? Need to perform balances around stage 1.

Relationships for stage 1
vapor V1, y1 TMB: L0 + V2 = L1 + V1 CMB: L0x0 + V2y2 = L1x1 + V1y1 EB: L0h0 + V2H2 = L1h1 + V1H1 VLE: K1(T1,P) = y1/x1 reflux L0, x0 distillate D, xD stage 1 V2 y2 L1 x1 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 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
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,y2 L1,x1 stage 2 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. 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, xR D, xD replace rectifying by enriching here and below – parallel with stripping stage j CMB: Vj+1yj+1 = Ljxj + DxD Vj+1,yj+1 Lj,xj CMO: yj+1 = (L/V) xj + (D/V) xD D = V - L yj+1 = (L/V) xj + (1 - L/V) xD F, xF B, xB Relates compositions of passing streams.

Lewis analysis of rectifying column
Assume CMO (Vj = Vj+1 = V; Lj = Lj-1 = L) Need specified xD; xD = y1 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) y=x VLE •xD = x0 (x0,y1) op. line 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
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 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 y=x VLE stage 1 (x1,y1)• •xD= x0 (x0,y1) op. line •(x1,y2) (x2,y2)• stage 2 1. Plot VLE line 2. Draw y=x line •(x2,y3) 3. Plot xD on y=x (x3,y3)• stage 3 4. Plot yint = xD (1 - L/V) •xB Lecture 7 ends here L/V = R/(R+1) = 2/3 yint = xD(1 - L/V)= 0.8/3 = 0.26 lowest xB possible for this op. line yint• 5. Draw in operating line 6. Step off stages from xD to xB 7. Count the stages N = 3

Limiting cases: rectification
L/V = 0 No reflux! Specifications: xD = 0.8, vary R = L/D y=x VLE L  0 R = L/D  0 NO REFLUX L/V  0 stage 1 (x1,y1)• •xD= x0 (x0,y1) Column operates like a single equilibrium stage. (Why bother?) 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 L/V = 1 Total reflux! Max. distance between VLE and op. line Max. separation on each equil. stage Corresponds to Nmin, but no distillate!

Minimum reflux ratio L/V = 0 Specifications: xD = 0.8, vary R VLE •
y=x VLE •xD= x0 (x0,y1) The number of stages N required to reach the VLE-op. line intersection point is . Increasing R = L/D Decreasing D Decreasing xB (for fixed N) This represents xB,min for a particular R. Rmin for this xB It also represents Rmin for this value of xB. xB ,min for this R L/V = 1 0 ≤ L/V ≤ 1 0 ≤ R ≤ 

Ractual can be specified as a multiple of Rmin
Optimum reflux ratio Rmin total cost capital cost Ropt operating (energy) cost cost/lb ∞ stages min. heat required 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.
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, xF D, xD Vk,yk Lk-1, xk-1 stage k CMB: Lk-1xk-1 = Vkyk + BxB CMO: yk = (L/V) xk-1 - (B/V) xB L = V + B yk = (L/V) xk-1 + (1 - L/V) xB B, xB

Graphical analysis of stripping column
equation of the operating line: y = (L/V) x + (1 - L/V) xB slope = L / V always positive y=x VLE op. line (xN+1,yN+1) PR plotting the operating line: y = x = (L/V) x + (1 - L/V) xB = xB plot xB on y = x finding 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)

McCabe-Thiele analysis: stripping column
Plot VLE line (yi vs. xi) Draw the y = x line Plot xB on y = x 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. (0.7,1) op. line Column with partial reboiler Specifications: xB = 0.07, Find N required to achieve xD = 0.55 y=x VLE stage 1 (xN-2,yN-2) stage 2 xD,max for this boilup ratio (xN-1,yN-1) • •(xN-2,yN-2) stage 3 xD 1. Plot VLE line (xN,yN) • • (xN-1,yN) 2. Draw y=x line 3. Plot xB = xN+1 on y = x 4. 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 = xD 7. Count the stages.

Limiting cases: stripping
Specifications: xB = 0.07, vary boilup ratio y=x VLE NO BOILUP Behaves as if the column wasn’t even there. (Why bother?) TOTAL BOILUP 2. B  0 Operating line is y=x TOTAL BOILUP PR NO BOILUP xB= xN+1 Max. distance between VLE and op. line Max. separation on each equil. stage Corresponds to Nmin. But no bottoms product!

Minimum boilup ratio Specifications: xB = 0.07, vary boilup ratio
yD ,max for this boilup ratio y=x VLE The number of stages N required to reach the VLE-op. line intersection point is . Increasing boilup ratio Increasing xD (for fixed N) Decreasing B Total boilup This represents yD,max for a particular boilup ratio. PR It also represents the minimum boilup ratio for this value of yD. No boilup xB= xN+1

McCabe-Thiele analysis of complete distillation column
Total condenser, partial reboiler Specifications: xD = 0.8, R = 2 xB = 0.07, Find N required Locate feed stage y=x bottom op. line Feed enters on stage 2 VLE stage 1 xB •xD top op. line stage 2 1. Draw y=x line 2. Plot xD and xB on y=x 3. Draw both op. lines PR 4. Step off stages starting at either end, using new op. line as you cross their intersection NEVER step over the VLE line. 5. Stop when you reach the other endpoint 6. Count stages 7. Identify feed stage

Feed condition Changing the feed temperature affects internal flow rates in the column V L feed F If the feed enters as a saturated liquid, the liquid flow rate below the feed stage will increase: V L If 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

Feed quality, q EB: rearrange: TMB: substitute: combine terms:
define: q  mol sat’d liquid generated on feed plate, per mol feed

Different types of feed quality
and saturated liquid feed q = 1 saturated vapor feed q = 0 feed flashes to form 2-phase < q < 1 mixture, q% liquid subcooled liquid feed q > 1 - some vapor condenses on feed plate superheated vapor q < 0 - some liquid vaporizes on feed plate

Equation of the feed line
rectifying section CMB: stripping section CMB: intersection of top and bottom operating lines: substitute: Wankat gives slope as q/(q-1) and equation of the feed line:

Plotting the feed line y = x = zF
where does the feed line intersect y=x? y=x VLE subcooled liq 2-phase sat'd liq y = x = zF sat'd vapor •zF Wankat gives slope as q/(q-1) superheated vap feed type q slope, m sat'd liquid q=1 m =  sat'd vapor q=0 m = 0 2-phase liq/vap 0<q<1 m < 0 subcooled liq q>1 m > 1 superheated vap 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. y=x 1 6 2 5 3 4 xB •xD •zF N = 6 + PR 1. Draw y=x line Operating lines intersect on stage 4. This is NF,opt. 2. Plot xD, xB and zF on y=x 3. Draw feed line, slope = -0.5 4. Draw top op. line, slope = L/V = 0.5 Optimum 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.

Feed lines in rectifying/stripping columns
rectifying column stripping column zF •xD • yD top operating line •zF PR bottom operating line May be slightly incompatible with previous description of these column endpoints - check •xB xB total condenser, no reboiler sat’d vapor feed, liquid bottoms F and B are passing streams 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: heat the feed • xB
•xD •zF Rmin xB •xD •zF pinch point decrease R qmin pinch point choice 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.

Another type of pinch point
Ethanol-water xD = 0.82, xB = 0.07 zF = 0.5, q = 0.5 Find Rmin y=x xB •xD •zF pinch point 1. Draw y=x line VLE 2. Plot xD, xB and zF on y=x 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.

Column with two feeds: bottoms B, xB distillate D, xD Column with three products: distillate D, xD bottoms B, xB feed F, z V L V L feed 2 F2, z2, q2 side-stream S, xS or yS z2 > z1 and/or q2 > q1 side-streams must be saturated liquid or vapor feed 1 F1, z1, q1 V L V L 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 y=x VLE 1 PR 2 5 3 4 xB •xD •z2 •z1 1. Draw y=x line 2. Plot xD, z1, z2 and xB on y=x 3. Draw both feed lines •z1 = z2 4. Draw top op. line, slope = L/V 5. 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 starting at either end, using new op. line each time you cross an intersection

Slope of middle operating line
2-feed mass balances: TMB: F2 + V´ = L´ + D CMB: F2z2 + V´yj+1 = L´xj + DxD feed 2 F2, z2, q2 D, xD middle operating line equation: y = (L´/V´)x + (DxD - F2z2)/V´ obtain slope from: L´ = F2q2 + L = F2q2 + (R)(D) V´ = L´ + D – F2 stage j side-stream  feed-stream with –ve flow rate sat’d liq y = x = xS sat’d vapor y = x = yS side-stream S, xS or yS D, xD Not finished 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´

McCabe-Thiele analysis of side-streams
Saturated liquid side-stream, xs = 0.64 Saturated vapor side-stream, ys = 0.73 VLE y=x y=x xB •xD •xS •z xB •xD •yS •z VLE Side-stream must correspond exactly to stage position.

Partial condensers y=x •
A partial condenser can be used when a vapor distillate is desired: PC 1 2 •yD D, yD V L L, x0 V, y1 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

Total reboilers A total reboiler is simpler (less expensive) than a partial reboiler and is used when the bottoms stream is readily vaporized: y=x TR N N-1 B, xB V L stage N V, yB •xB,yB A total reboiler is not an equilibrium stage.

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: Can also define Murphree liquid efficiency: where yn* is the equilibrium vapor composition (not actually achieved) on stage n: yn* = Kn xn 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. y=x xB •xD •z 2 8 3 6 4 5 1 7 1. Draw y=x line 2. Plot xD, z, and xB on y=x 3. Draw feed line 4. Draw top op. line, slope = L/V 5. Draw bottom operating line (no calc. required) N = 8 + PR PR NF,opt = 6 6. Find partial reboiler 7. Step off stages, using EMV to adjust vertical step size. 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: 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 by 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 condenser distillate D, xD bottoms B, xB feed F, z V L S, xS intermediate reboiler yS = xS V´´ L´´ V L Each column section has its own operating line.

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. V2 L1 L0, x0 V1, y1 D, xD stage 1 c 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 L1 = L0 + c CMO is valid below stage 1. Find L/V = L1/V2? 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 q0  quality of reflux Subcooled reflux causes L/V to increase. Superheated boilup causes L/V to increase. where L0/V1 = (L0/D)/(1 + L0/D) = R/(R + 1)

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 Vj+1, yj+1 Lj, xj TMB: V + B = L + S usually 0 CMB: V yj+1 + B xB = L xj + S yS mostly H2O bottoms B, xB CMO: B = L S, yS B, xB 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. y=x VLE 1 6 2 5 3 4 •xD •zF 1. Draw y=x line 2. Plot xD and zF on y=x 3. Plot xB on x-axis 4. Draw feed line, slope = q/(q-1) 5. Draw top op. line, slope = L/V All stages are on the column (no partial reboiler). 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 N = 6 NF,opt = 4 xB

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 Bubble cap 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 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) vertical downcomer alternates sides Dual-pass tray Add pipe downcomer four-pass tray

Tray efficiency design point excessive entrainment efficiency inefficient mass transfer flooding weeping/dumping 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: Flooding: Weeping:

Column flooding 1. jet flood (due to entrainment)
3. insufficient downcomer clearance • weir height above downcomer • liquid backs up downcomer 2. 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.

Column sizing 1. Calculate vapor flood velocity, uflood (ft/s)
where Csb,f is the capacity factor, from empirical correlation with flow parameter, FP 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 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: 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 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)