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Batch Distillation

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**History Early distillation of alcohol**

Alembic still for distillation of brandy

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**Major types of batch distillation**

Simple batch distillation Multistage batch distillation

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**Reasons to use batch distillation**

Small capacity (e.g., specialty chemicals) Intermittent need Test run for a new product Up-stream operations are batch (e.g., alcoholic spirits) Feed inappropriate for continuous distillation (suspended solids) Feed varies widely in composition

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**Simple batch distillation no rectification ( = no column)**

Characteristics: no column; a single equilibrium stage (= the still pot) single charge (F) to still pot at time = 0 vapor is withdrawn continuously composition of liquid in still pot (xW) changes continuously composition of liquid distillate (xD) changes continuously time t: W, xW V, y D, xD time 0: F, xF still pot with heater

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**Rayleigh equation TMB: F = Wfinal + Dtotal**

time t: W, xW y = xD V, y D, xD TMB: F = Wfinal + Dtotal CMB: FxF = WfinalxW,final + DtotalxD,avg Specify F, xF and xW,final or xD,avg Leaves 3 unknowns: Wfinal, Dtotal and xW,final or xD,avg Need one more equation: time 0: F, xF still pot, with heater dCMB: - xDdW = - d(WxW) (vapor withdrawn) = (change in still pot composition) chain rule: - xDdW = - WdxW - xWdW WAIT! K is not constant; K = K(T) Rayleigh equation where xD = f(xW)

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**Integration of the Rayleigh equation**

1/(xD – xD) x = xW Numerical integration: Constant relative volatility: xW,final • • xF Specify F, xF, and either Wfinal or xW,final. Simpson’s rule:

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**Solvent switching using simple batch distillation**

Goal: replace one solvent by another, in order to facilitate crystallization of a non-volatile product, or for a subsequent reaction step. The Hard Way: Boil off most of original solvent in a batch still. Add second solvent. Perform second batch distillation to remove residual original solvent. The Easy Way: Constant-level batch distillation Add second solvent continuously as first solvent vaporizes, keeping W constant; more energy efficient and uses less solvent. dTMB: dV = dS (vapor withdrawn) = (new solvent added) dCMB: ydV = - xDdS = WdxW (W constant)

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**Batch steam distillation**

Used for thermally fragile organics (e.g., essential oils in perfume industry), and for slurries/sludges containing organics. W, xW V, y D, xD = 1 A single charge (F) added to still pot at time = 0. Steam is added continuously. H2O(l) (to waste) still pot, no heater steam If W, D are immiscible with water, we have a heterogeneous azeotrope. H2O(l) D.o.F. = 2 components – 3 phases + 2 = 1 Fix Ptotal, then T cannot vary! constant T < Tbp(H2O) How much steam is required? Both H2O and organic vaporize well below their single-component boiling points. Also, constant vapor composition. Raoult’s Law: Ptotal = P*WxW + P*H2O Steam also needed to heat and vaporize the material in the still pot.

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**Batch distillation with rectification**

TMB: Vj+1 = Lj + D CMB: Vj+1yj+1 = Ljxj + DxD • both are time-dependent • either D or xD (or both) change over the course of the distillation time t: W, xW y1 = x0 = xD V1, y1 D, xD L0, x0 stage 1 y1 ≠ K / xW stage j CMO: Vj+1 = Vj and Lj = Lj-1 operating line equation: y = (L/V) x + (1 - (L/V)) xD y = x = xD slope = L/V • actually a family of operating lines, since L/V or xD (or both) change over the course of the distillation • therefore the operating line moves on the M-T diagram Vj+1, yj+1 Lj, xj stage N VN+1, xN+1 LN, xN time 0: F, xF still pot, with heater stage N+1

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**Choice of operating methods**

Constant reflux ratio (variable xD) y=x Constant distillate composition (variable R) VLE y=x •xD •xD total reflux • time VLE time • • • distillation must end when (or before) xD,avg = xF distillation must end when (or before) R = ∞ (L/V = 1) Easy to monitor and control. Harder to monitor and control (need to detect xD on-stream and adjust R accordingly). Can solve graphically, if we assume no liquid holdup on the column.

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**Multistage batch distillation with constant R**

Given F, xF, xW,final, R and N, find Dtotal, xD,avg y=x VLE 1 2 For N = 2 (incl. reboiler) • xD,4 •xD,1 •xD,2 •xD,3 • • 1. For an arbitrary set of xD values, draw a series of parallel operating lines, each with slope R/(R+1) • 1 2 xW,2 • 1 2 xW,3 2. Step off N stages on each operating line to find its corresponding xW 3. Perform numerical integration: plot 1/(xD-xW) vs xW limits: xF, xW,final • 1 2 xW,4 4. Calculate Wfinal using Rayleigh equation xW,1 5. Solve mass balances for Dtotal and xD,avg If xD,avg is specified instead of xW,final: guess xW,final, calculate xD,avg, iterate.

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**Operating time at constant R (D)**

depends on vapor flow rate (V), which depends on boilup rate shut down, cleaning and recharging still pot, restart • if the boilup rate is constant, then V is constant, and D will be constant condenser TMB: • V = Vmax when vapor velocity u = uflood • uflood depends on column diameter • typically, operate at D = 0.75 Dmax

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**Calculating column diameter**

We want to use the smallest diameter that will not cause the column to flood. where σ is surface tension, ρL and ρV are liquid and vapor densities, respectively. Csb,flood is the capacity factor, depends on flow parameter FP and tray spacing; obtain from graphical correlation. where η is the fraction of the column cross-sectional area available for vapor flow (i.e., column cross-sectional area minus downcomer area).

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**Multistage batch distillation with constant xD**

Given F, xF, xD (maybe) xw,final and N, find Rinitial, Rmin, xW,min y=x VLE N = ∞, R = Rmin • •xD (L/V)initial (L/V)min 1. Draw trial op. lines and step off N stages to end at xF This is trial-and-error, except for N = 2, or N = ∞ (Rmin) xF • • xW.min 3. Find xW,min using (L/V) = 1. Rayleigh equation not needed! Verify xW,final > xW,min. Is xWfinal required? 4. Solve mass balance for Wfinal and Dtotal. N = 2 (incl. reboiler)

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**Operating time with constant xD**

mass balance: x = xW Numerical integration: xW,final • xF 1. Draw a series of arbitrary operating lines, each with a different slope L/V 2. Step off N stages on each operating line to find its corresponding xW 3. Perform numerical integration (plot graph, use Simpson’s rule) 4. Calculate toperating

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**Effect of liquid holdup on the column**

Optimal control • use optimal, time-dependent reflux ratio (not constant R, not constant xD) • more energy-efficient • useful for difficult separations Effect of liquid holdup on the column • usually, we can assume vapor holdup is negligible • liquid holdup causes xw to be lower than it would be in the absence of holdup • causes the degree of separation to decrease To assess the effect on batch distillation: • measure the amount of holdup at total reflux • perform computational simulation Next year: add heater and condenser duties

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© 2015 Carl Lund, all rights reserved A First Course on Kinetics and Reaction Engineering Class 31.

© 2015 Carl Lund, all rights reserved A First Course on Kinetics and Reaction Engineering Class 31.

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