Batch Distillation

History Early distillation of alcohol Alembic still for distillation of brandy http://essentialspirits.com/history.htm

Major types of batch distillation Simple batch distillation Multistage batch distillation

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

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no column; a single equilibrium stage (= the still pot) single charge (F) to still pot at time = 0 vapor is withdrawn continuously composition of liquid in still pot (xW) changes continuously composition of liquid distillate (xD) changes continuously time t: W, xW V, y D, xD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

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and xW,final or xD,avg Need one more equation: time 0: F, xF still pot, with heater dCMB: - xDdW = - d(WxW) (vapor withdrawn) = (change in still pot composition) chain rule: - xDdW = - WdxW - xWdW WAIT!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)

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:

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 dTMB: dV = dS (vapor withdrawn) = (new solvent added) dCMB: 0 - ydV = - xDdS = WdxW (W constant)

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.

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

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.

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.

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

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).

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)

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

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