Presentation on theme: "Continuous Column Distillation. Column diagram feed F, x F bottoms B, x B boilup V, y B reflux L, x R distillate D, x D feed stage total condenser reflux."— Presentation transcript:
Continuous Column Distillation
Column diagram feed F, x F bottoms B, x B boilup V, y B reflux L, x R distillate D, x D feed stage total condenser reflux drum (accumulator) VL liquid/vapor streams inside the column flow counter- current in direct contact with each other partial reboiler x R = x D y B x B x D K x B temperature all three external streams (F, D, B) can be liquids (usual case) for a binary mixture, the compositions x F, x D, x B all refer to the more volatile component 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 enriching section stripping section
External mass balance TMB:F = D + B CMB:F x F = D x D + B x B for specified F, x F, x D, x B, there are only 2 unknowns (D, B) B = F - D feed F, x F bottoms B, x B distillate D, x D
External energy balance assume column is well- insulated, adiabatic feed F, x F bottoms B, x B distillate D, x D EB:F h F + Q C + Q R = D h D + B h B F, h F are known D and B are saturated liquids so h D, h B are also known unknowns: Q C, Q R need another equation
Balance on condenser distillate D, x D reflux L 0, x R vapor V 1, y 1 1. Mass balance TMB:V 1 = D + L 0 CMB:y 1 = x D = x R (doesnt help) unknowns: V 1, L 0 specify external reflux ratio R = L 0 /D V 1 = D + (L 0 /D)D = (1 + R)D 2. Energy balance V 1 H 1 + Q C = (D + L 0 )h D = V 1 h D Q C = V 1 (h D – H 1 ) then calculate Q R from column energy balance h D > H 1 Q C < 0 Q R > 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: x F = 0.46
Calculating fractional recoveries B = F – D = = 339
vapor V 1, y 1 distillate D, x D Stage-by-stage analysis Lewis-Sorel method reflux L 0, x 0 V2y2V2y2 L1x1L1x1 stage 1 Consider the top of the distillation column: V 1, V 2 are saturated vapors L 0, L 1 are saturated liquids Which streams have compositions related by VLE? V 1, L 1 They are streams leaving the same equilibrium stage. K 1 (T 1,P) = y 1 /x 1 How are the compositions of streams V 2 and L 1 related? Need to perform balances around stage 1.
Relationships for stage 1 TMB:L 0 + V 2 = L 1 + V 1 CMB:L 0 x 0 + V 2 y 2 = L 1 x 1 + V 1 y 1 EB: L 0 h 0 + V 2 H 2 = L 1 h 1 + V 1 H 1 VLE:K 1 (T 1,P) = y 1 /x 1 vapor V 1, y 1 distillate D, x D reflux L 0, x 0 V2y2V2y2 L1x1L1x1 stage 1 There are 14 variables: 4 flow rates (L 1, V 2, L 0, V 1 ) 4 compositions (x 1, y 2, x 0, y 1 ) 4 enthalpies (h 1, H 2, h 0, H 1 ) T 1, P We usually specify 10 of them: P, x D, D, R = L 0 /D x D = x 0 = y 1 V 1 = L 0 + D T 1 and all 4 enthalpies (by VLE) 4 unknowns (L 1, x 1, V 2, y 2 ) and 4 equations: problem is completely specified.
Relationships for stage 2 TMB:L 1 + V 3 = L 2 + V 2 CMB:L 1 x 1 + V 3 y 3 = L 2 x 2 + V 2 y 2 EB: L 1 h 1 + V 3 H 3 = L 2 h 2 + V 2 H 2 VLE:K 2 (T 2,P) = y 2 /x 2 can solve for 4 unknowns (L 2, x 2, V 3, y 3 ) V 2,y 2 L 1,x 1 and so on… proceed down the column to the reboiler. Very tedious. Simplifying assumption: If i (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 stage 2 V 3,y 3 L 2,x 2
Constant molal overflow TMB: L 1 + V 3 = L 2 + V 2 CMO: V 3 - V 2 = L 2 - L 1 = 0 V 3 = V 2 = V L 2 = L 1 = 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 B, x B D, x D F, x F 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, x R stage j V j+1,y j+1 L j,x j CMB: V j+1 y j+1 = L j x j + Dx D CMO:y j+1 = (L/V) x j + (D/V) x D D = V - L y j+1 = (L/V) x j + (1 - L/V) x D Relates compositions of passing streams.
Lewis analysis of rectifying column 1.Assume CMO (V j = V j+1 = V; L j = L j-1 = L) 2.Need specified x D ; x D = y 1 3.Stage 1: use VLE to obtain x 1 x 1 = y 1 /K 1 (T 1,P) 4.Use mass balance to obtain y 2 y 2 = (L/V) x 1 + (1 - L/V) x D 5.Stage 2: use VLE to obtain x 2 x 2 = y 2 /K 2 (T 2,P) 6.Use mass balance to obtain y 3 y 3 = (L/V) x 2 + (1 - L/V) x D 7.Continue until x = x B
Graphical analysis of rectifying column equation of the operating line: y = (L/V) x + (1 - L/V) x D slope = (L/V) always positive (compare to flash drum) plotting the operating line: y int = (1 - L/V) x D recall: x D = x R = x 0 ; the passing stream is y 1 the operating line starts at the point (x 0,y 1 ) the operating line gives the compositions of all passing streams (x j,y j+1 ) find a second point on the operating line: y = x = (L/V) x + (1 - L/V) x D = x D plot x D on y = x VLE y=x y int op. line x D = x 0 (x 0,y 1 )
McCabe-Thiele analysis: rectifying column 1.Plot VLE line (y i vs. x i ) 2.Draw the y = x line 3.Plot x D on y = x 4.Plot y int = x D (1 – L/V) L/V internal reflux ratio, usually not specified instead, the external reflux ratio (R) is specified 5.Draw in the operating line 6.Step off stages, alternating between VLE and operating line, starting at (x 0,y 1 ) located at y = x = x D, until you reach x = x B 7.Count the stages.
Ex.: MeOH-H 2 O rectifying column VLE y=x y int op. line (x 1,y 1 ) (x 2,y 2 ) (x 3,y 3 ) (x 1,y 2 ) (x 2,y 3 ) xBxB 2. Draw y=x line 3. Plot x D on y=x 4. Plot y int = x D (1 - L/V) 6. Step off stages from x D to x B 7. Count the stages Rectifying column with total condenser Specifications: x D = 0.8, R = 2 Find N required to achieve x B = 0.1 L/V = R/(R+1) = 2/3 y int = x D (1 - L/V)= 0.8/3 = 0.26 NEVER step over the VLE line. lowest x B possible for this op. line 1. Plot VLE line 5. Draw in operating line x D = x 0 (x 0,y 1 ) stage 1 stage 2 stage 3 N = 3
Limiting cases: rectification x D = x 0 (x 0,y 1 ) VLE y=x stage 1 (x 1,y 1 ) Specifications: x D = 0.8, vary R = L/D 1.L 0 R = L/D 0 NO REFLUX L/V 0 L/V = 0 No reflux! Max. distance between VLE and op. line Max. separation on each equil. stage Corresponds to N min, but no distillate! L/V = 1 Total reflux! 2.D 0 R = L/D TOTAL REFLUX L/V = R/(R+1) 1 (LHôpitals Rule) Operating line is y=x Column operates like a single equilibrium stage. (Why bother?) 0 L/V 1 0 R
Minimum reflux ratio VLE y=x Specifications: x D = 0.8, vary R L/V = 0 The number of stages N required to reach the VLE-op. line intersection point is. L/V = 1 0 L/V 1 0 R x B,min for this R This represents x B,min for a particular R. It also represents R min for this value of x B. x D = x 0 (x 0,y 1 ) R min for this x B Increasing R = L/D Decreasing D Decreasing x B (for fixed N)
Optimum reflux ratio cost/lb external reflux ratio, R R min R opt operating (energy) cost Rule-of-thumb: 1.05 R opt /R min 1.25 R actual can be specified as a multiple of R min stages min. heat required capital cost total cost
Stripping column B, x B D, x D F, x F 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: stage k CMB: L k-1 x k-1 = V k y k + Bx B V k,y k L k-1, x k-1 CMO:y k = (L/V) x k-1 - (B/V) x B L = V + B y k = (L/V) x k-1 + (1 - L/V) x B
Graphical analysis of stripping column Where is the partial reboiler? Designate this as stage N+1, with x N+1 = x B. VLE y=x op. line x B = x N (x N+1,y N+2 ) equation of the operating line: y = (L/V) x + (1 - L/V) x B slope = L / V always positive plotting the operating line: y = x = (L/V) x + (1 - L/V) x B = x B plot x B on y = x finding the operating line slope: (recall V/B is the boilup ratio) Coordinates of the reboiler: (x N+1,y N+1 ) (x N+1,y N+1 ) PR
McCabe-Thiele analysis: stripping column 1.Plot VLE line (y i vs. x i ) 2.Draw the y = x line 3.Plot x B on y = x 4.Draw in the operating line 5.Step off stages, alternating between VLE and operating line, starting at (x N+1,y N+2 ) located at y = x = x B, until you reach x = x D 6.Count the stages.
Ex.: MeOH-H 2 O stripping column VLE y=x (0.7,1) op. line (x N-2,y N-2 ) stage 2 2. Draw y=x line 3. Plot x B = x N+1 on y = x 4. Draw op. line 5. Step off stages starting at PR 6. Stop when you reach x = x D Column with partial reboiler Specifications: x B = 0.07, Find N required to achieve x D = Count the stages. (x N+1, y N+1 ) (x N,y N+1 ) PR (x N,y N ) (x N-1,y N ) (x N-1,y N-1 ) x D,max for this boilup ratio stage 3 NEVER step over the VLE line. (x N-2,y N-2 ) stage 1 x B = x N+1 (x N+1,y N+2 ) 1. Plot VLE line xDxD
Limiting cases: stripping VLE y=x Specifications: x B = 0.07, vary boilup ratio Max. distance between VLE and op. line Max. separation on each equil. stage Corresponds to N min. But no bottoms product! Behaves as if the column wasnt even there. (Why bother?) x B = x N+1 PR 1. NO BOILUP 2.B 0 Operating line is y=x TOTAL BOILUP NO BOILUP TOTAL BOILUP
Minimum boilup ratio VLE y=x Specifications: x B = 0.07, vary boilup ratio The number of stages N required to reach the VLE-op. line intersection point is. Increasing boilup ratio Decreasing B Increasing x D (for fixed N) y D,max for this boilup ratio This represents y D,max for a particular boilup ratio. It also represents the minimum boilup ratio for this value of y D. x B = x N+1 PR No boilup Total boilup
McCabe-Thiele analysis of complete distillation column VLE y=x stage 1 1. Draw y=x line 2. Plot x D and x B 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 Total condenser, partial reboiler Specifications: x D = 0.8, R = 2 x B = 0.07, Find N required Locate feed stage 6. Count stages PR stage 2 x B xDxD top op. line bottom op. line Feed enters on stage 2 NEVER step over the VLE line. 7. Identify feed stage
If the feed enters as a saturated liquid, the liquid flow rate below the feed stage will increase: If the feed enters as a saturated vapor, the vapor flow rate above the feed stage will increase: Feed condition Changing the feed temperature affects internal flow rates in the column VL feed F VL and 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:
Different types of feed quality saturated liquid feed q = 1 feed flashes to form 2-phase0 < q < 1 mixture, q% liquid and subcooled liquid feedq > 1 - some vapor condenses on feed plate superheated vapor q < 0 - some liquid vaporizes on feed plate saturated vapor feedq = 0
Equation of the feed line rectifying section CMB: and stripping section CMB: intersection of top and bottom operating lines: substitute: equation of the feed line:
Plotting the feed line where does the feed line intersect y=x? y=x VLE y = x = z F feed type q slope, m sat'd liquid q=1 m = sat'd vapor q=0m = 0 2-phase liq/vap 01m > 1 superheated vap q<00
"name": "Plotting the feed line where does the feed line intersect y=x.",
"description": "y=x VLE y = x = z F feed type q slope, m sat d liquid q=1 m = sat d vapor q=0m = 0 2-phase liq/vap 01m > 1 superheated vap q<00
Ex.: Complete MeOH-H 2 O column y=x 1. Draw y=x line 2. Plot x D, x B and z F on y=x 3. Draw feed line, slope = 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. Total condenser, partial reboiler Specifications: x D = 0.9, x B = 0.04, z F = 0.5, R=1 Feed is a 2-phase mixture, 50% liq. Find N and N F,opt. 4. Draw top op. line, slope = L/V = 0.5 x B xDxD zFzF Operating lines intersect on stage 4. This is N F,opt. Using a non-optimal feed location reduces separation. N = 6 + PR PR We can independently specify only 2 of the following 3 variables: R, q, V/B (usually: R, q).
Feed lines in rectifying/stripping columns PR x B zFzF zFzF xDxD stripping column partial reboiler, no condenser satd liquid feed, vapor distillate F and V are passing streams rectifying column total condenser, no reboiler satd vapor feed, liquid bottoms F and B are passing streams xBxB bottom operating line top operating line y D
Design freedom choice of R dictates required boilup ratio. Fixed q. Vary R: x B xDxD zFzF Fixed R. Vary q: x B xDxD zFzF R min heat the feed q min pinch point decrease R pinch point 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 VLE y=x 1. Draw y=x line 2. Plot x D, x B and z F on y=x 3. Draw feed line, slope = q/(q-1) 5. Dont cross the VLE line! Ethanol-water x D = 0.82, x B = 0.07 z F = 0.5, q = 0.5 Find R min 4. Draw top op. line to intersect with feed line on VLE line x B xDxD zFzF 6. Redraw top operating line as tangent to VLE. pinch point
Additional column inputs/outputs bottoms B, x B distillate D, x D VL feed 2 F 2, z 2, q 2 V´L´ feed 1 F 1, z 1, q 1 VL distillate D, x D bottoms B, x B feed F, z VL side-stream S, x S or y S V´L´ VL Column with two feeds: Column with three products: Each intermediate input/output stream changes the mass balance, requiring a new operating line. z 2 > z 1 and/or q 2 > q 1 side-streams must be saturated liquid or vapor
Multiple feedstreams VLE y=x 1. Draw y=x line 2. Plot x D, z 1, z 2 and x B on y=x 3. Draw both feed lines 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 4. Draw top op. line, slope = L/V 1 PR Optimum location for feed 1 is stage 5. Optimum location for feed 2 is stage 3. Total condenser, partial reboiler Specifications: x D = 0.9, x B = 0.07, z 1 = 0.4, z 2 =0.6 Some specified q-values R = 1. Find N, N F1,opt, N F2,opt x B xDxD z2z2 z1z1 5. Calculate slope of middle operating line, L´/V´, and draw middle operating line z 1 = z 2
Slope of middle operating line 2-feed mass balances: TMB: F 2 + V´ = L´ + D CMB: F 2 z 2 + V´y j+1 = L´x j + Dx D middle operating line equation: y = (L´/V´)x + (Dx D - F 2 z 2 )/V´ obtain slope from: L´ = F 2 q 2 + L = F 2 q 2 + (R)(D) V´ = L´ + D – F 2 feed 2 F 2, z 2, q 2 D, x D V´L´ stage j side-stream S, x S or y S D, x D V´L´ stage j middle operating line equation: y = (L´/V´)x + (Dx D + Sx S )/V´ side-stream feed-stream with –ve flow rate satd liq y = x = x S satd vapory = x = y S side-stream mass balances: TMB: V´- L´= D + S CMB: V´y j+1 - L´x j = Dx D + Sx S
McCabe-Thiele analysis of side-streams VLE y=x Saturated liquid side-stream, x s = 0.64 xB xB xDxD xSxS z Saturated vapor side-stream, y s = 0.73 VLE y=x xB xB xDxD ySyS z Side-stream must correspond exactly to stage position.
Partial condensers A partial condenser can be used when a vapor distillate is desired: D, y D VL L, x 0 V, y 1 A partial condenser is an equilibrium stage. CMB: Vy j+1 = Lx j + Dy D Operating line equation: y = (L/V)x + Dy D = (L/V)x + (1 - L/V)y D y=x yDyD PC 1 2
Total reboilers A total reboiler is simpler (less expensive) than a partial reboiler and is used when the bottoms stream is readily vaporized: A total reboiler is not an equilibrium stage. y=x x B,y B TR N N-1 B, x B VL stage N V, y B
Stage efficiency Under real operating conditions, equilibrium is approached but not achieved: N actual > N equil overall column efficiency:E overall = N equil /N actual Efficiency can vary from stage to stage. Reboiler efficiency tray efficiency where y n * is the equilibrium vapor composition (not actually achieved) on stage n: y n * = K n x n Can also define Murphree liquid efficiency: x j * = y j / K j Murphree vapor efficiency:
y=x x B xDxD z PR Ex.: Vapor efficiency of MeOH-H 2 O column Total condenser, partial reboiler Specifications: x D = 0.9, x B = 0.07, z = 0.5, q = 0.5, R = 1, E MV,PR = 1, E MV = Find N and N F,opt. 1. Draw y=x line 2. Plot x D, z, and x B on y=x 3. Draw feed line 5. Draw bottom operating line (no calc. required) 6. Find partial reboiler 4. Draw top op. line, slope = L/V 7. Step off stages, using E MV to adjust vertical step size. 8. Label real stages N = 8 + PR N F,opt = 6 To use E LV, 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 distillate D, x D bottoms B, x B feed F, z VL VL V´L´ S, x S intermediate reboiler y S = x S Each column section has its own operating line. V´´L´´
V2V2 L1L1 L 0, x 0 V 1, y 1 D, x D stage 1 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. EB: V 2 H 2 + L 0 h 0 = V 1 H 1 + L 1 h 1 where H 1 H 2 = H, but h 0 h 1 = h (V 2 – V 1 )H = L 1 h - L 0 h 0 cH = (L 0 + c)h - L 0 h 0 = L 0 (h - h 0 ) + ch where L 0 /V 1 = (L 0 /D)/(1 + L 0 /D) = R/(R + 1) CMO is valid below stage 1. Find L/V = L 1 /V 2 ? V 1 = V 2 - c and L 1 = L 0 + c c Pumping a saturated liquid damages the pump, by causing cavitation. The reflux stream (L 0 ) should be subcooled. This will cause some vapor to condense. Subcooled reflux causes L/V to increase. Superheated boilup causes L/V to increase. q 0 quality of reflux
Open steam distillation B, x B D, x D mostly MeOH S, y S L, x R stage j V j+1, y j+1 Lj,xjLj,xj MeOH/H 2 O feed F, z bottoms B, x B If the bottoms stream is primarily water, then the boilup is primarily steam. Can replace reboiler with direct steam heating (S). Top operating line and feed lines do not change. Bottom operating line is different: TMB:V + B = L + S CMB:V y j+1 + B x B = L x j + S y S usually 0 Operating line equation: y = (L/V) x - (L/V) x B x int : x = x B CMO:B = L mostly H 2 O
Ex.: Open steam distillation of MeOH/H 2 O VLE y=x 1. Draw y=x line 2. Plot x D and z F on y=x 4. Draw feed line, slope = q/(q-1) 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 5. Draw top op. line, slope = L/V x B xDxD zFzF 3. Plot x B on x-axis All stages are on the column (no partial reboiler). N = 6N F,opt = 4 Specifications: x D = 0.9, x B = 0.07, z F = 0.5 Feed is a 2-phase mixture, 50% liq. Total condenser, open steam, R = 1. Find N and N F,opt.
Column internals 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 Sieve tray
Other types of trays 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 Valve 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 Bubble cap tray
Downcomers Dual-pass tray In large diameter columns, use multi-pass trays to reduce liquid loading in downcomers Cross-flow tray (single pass) vertical downcomer alternates sides Both liquid and vapor pass through holes Narrow operating range Dual-flow tray (no downcomer)
Tray efficiency efficiency vapor flow rate weeping/du mping flooding inefficient mass transfer excessive entrainment design point 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) vapor flow rate too high 2. lack of downcomer seal weir height below downcomer vapor flows up downcomer 3. insufficient downcomer clearance weir height above downcomer liquid backs up downcomer ensure bottom edge of downcomer is 12´´ below top edge of outlet weir.
Column sizing 1. Calculate vapor flood velocity, u flood (ft/s) where C sb,f is the capacity factor, from empirical correlation with flow parameter, FP where W L and W V are the mass flow rates of liquid and vapor, respectively 2. Determine net area required for vapor flow, A net, based on operating vapor velocity, u op, ft/s where V is molar vapor flow rate and MW V is average molecular weight of vapor
Column sizing, cont. Relationship between net area for vapor flow, A net, in ft 2, and column diameter, D, in ft: where 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 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) structured packing:random packing: