Download presentation

Presentation is loading. Please wait.

Published byAllison Bennington Modified over 2 years ago

1
Batch Distillation

2
Early distillation of alcohol Alembic still for distillation of brandy History

3
Simple batch distillationMultistage batch distillation Major types of batch distillation

4
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

5
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-no column; a single equilibrium stage (= the still pot) -single charge (F) to still pot at time = 0 -vapor is withdrawn continuously -composition of liquid in still pot (x W ) changes continuously -composition of liquid distillate (x D ) changes continuously time t: W, x W V, y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

6
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Leaves 3 unknowns: W final, D total and x W,final or x D,avg Need one more equation: dCMB: - x D dW = - d(Wx W ) (vapor withdrawn) = (change in still pot composition) chain rule:- x D dW = - Wdx W - x W dW Rayleigh equation where x D = f(x W ) WAIT!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)

7
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

8
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 dTMB: dV = dS (vapor withdrawn) = (new solvent added) dCMB: 0 - ydV = - x D dS = Wdx W (W constant)

9
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.

10
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

11
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.

12
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

13
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:

14
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).

15
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

16
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

17
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

Similar presentations

© 2017 SlidePlayer.com Inc.

All rights reserved.

Ads by Google