Presentation on theme: "Batch Distillation. Early distillation of alcohol Alembic still for distillation of brandy History."— Presentation transcript:
Early distillation of alcohol Alembic still for distillation of brandy History
Simple batch distillationMultistage batch distillation Major types of batch distillation
Reasons to use batch distillation 1.Small capacity (e.g., specialty chemicals) 2.Intermittent need 3.Test run for a new product 4.Up-stream operations are batch (e.g., alcoholic spirits) 5.Feed inappropriate for continuous distillation (suspended solids) 6.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-no column; a single equilibrium stage (= the still pot) -single charge (F) to still pot at time = 0 -vapor is withdrawn continuously -composition of liquid in still pot (x W ) changes continuously -composition of liquid distillate (x D ) changes continuously time t: W, x W V, y D, x D time 0: F, x F still pot with heater
Rayleigh equation TMB: F = W final + D total time t: W, x W y = x D V, y D, x D time 0: F, x F still pot, with heater CMB: Fx F = W final x W,final + D total x D,avg Specify F, x F and x W,final or x D,avg Leaves 3 unknowns: W final, D total and x W,final or x D,avg Need one more equation: dCMB: - x D dW = - d(Wx W ) (vapor withdrawn) = (change in still pot composition) chain rule:- x D dW = - Wdx W - x W dW Rayleigh equation where x D = f(x W ) WAIT! K is not constant; K = K(T)
Integration of the Rayleigh equation Constant relative volatility: Simpsons rule: 1/(x D – x D ) x = x W Numerical integration: Specify F, x F, and either W final or x W,final. x W,final xFxF
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: 1.Boil off most of original solvent in a batch still. 2.Add second solvent. 3.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 dTMB: dV = dS (vapor withdrawn) = (new solvent added) dCMB: 0 - ydV = - x D dS = Wdx W (W constant)
Batch steam distillation W, x W V, y D, x D = 1 H 2 O(l) still pot, no heater steam Used for thermally fragile organics (e.g., essential oils in perfume industry), and for slurries/sludges containing organics. H 2 O(l) (to waste) How much steam is required? Steam also needed to heat and vaporize the material in the still pot. Also, constant vapor composition. Raoults Law: P total = P* W x W + P* H2O Both H 2 O and organic vaporize well below their single-component boiling points. Fix P total, then T cannot vary! constant T < T bp (H 2 O) D.o.F. = 2 components – 3 phases + 2 = 1 If W, D are immiscible with water, we have a heterogeneous azeotrope. A single charge (F) added to still pot at time = 0. Steam is added continuously.
Batch distillation with rectification TMB: V j+1 = L j + D CMB: V j+1 y j+1 = L j x j + Dx D both are time-dependent either D or x D (or both) change over the course of the distillation time t: W, x W y 1 = x 0 = x D V 1, y 1 D, x D time 0: F, x F still pot, with heater stage N+1 L 0, x 0 L N, x N V N+1, x N+1 stage 1 y 1 K / x W stage N V j+1, y j+1 L j, x j stage j CMO:V j+1 = V j and L j = L j-1 operating line equation: y = (L/V) x + (1 - (L/V)) x D y = x = x D slope = L/V actually a family of operating lines, since L/V or x D (or both) change over the course of the distillation therefore the operating line moves on the M-T diagram
Choice of operating methods VLE y=x Constant reflux ratio (variable x D ) y=x Constant distillate composition (variable R) VLE xDxD xDxD time total reflux distillation must end when (or before) x D,avg = x F distillation must end when (or before) R = (L/V = 1) Easy to monitor and control. Harder to monitor and control (need to detect x D on-stream and adjust R accordingly). Can solve graphically, if we assume no liquid holdup on the column.
Multistage batch distillation with constant R VLE y=x 1. For an arbitrary set of x D values, draw a series of parallel operating lines, each with slope R/(R+1) 2. Step off N stages on each operating line to find its corresponding x W 3. Perform numerical integration: plot 1/(x D -x W ) vs x W limits: x F, x W,final 4. Calculate W final using Rayleigh equation 1 2 For N = 2 (incl. reboiler) Given F, x F, x W,final, R and N, find D total, x D,avg x D,4 x D,1 x D,2 x D,3 5. Solve mass balances for D total and x D,avg If x D,avg is specified instead of x W,final : guess x W,final, calculate x D,avg, iterate. x W,1 1 2 x W,2 1 2 x W,3 1 2 x W,4
Operating time at constant R (D) V = V max when vapor velocity u = u flood u flood depends on column diameter typically, operate at D = 0.75 D max 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:
Calculating column diameter We want to use the smallest diameter that will not cause the column to flood. where σ is surface tension, L and ρ V are liquid and vapor densities, respectively. C sb,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 x D VLE y=x 1. Draw trial op. lines and step off N stages to end at x F This is trial-and-error, except for N = 2, or N = (R min ) 3. Find x W,min using (L/V) = 1. Rayleigh equation not needed! Verify x W,final > x W,min. N = 2 (incl. reboiler) xDxD 4. Solve mass balance for W final and D total. Given F, x F, x D (maybe) x w,final and N, find R initial, R min, x W,min (L/V) min N =, R = R min (L/V) initial x F x W.min
Operating time with constant x D 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 x W 3. Perform numerical integration (plot graph, use Simpsons rule) 4. Calculate t operating mass balance: x = x W Numerical integration: x W,final xFxF
Optimal control usually, we can assume vapor holdup is negligible liquid holdup causes x w 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 Effect of liquid holdup on the column use optimal, time-dependent reflux ratio (not constant R, not constant x D ) more energy-efficient useful for difficult separations