Batch Distillation

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Early distillation of alcohol Alembic still for distillation of brandy

History

Simple batch distillation Multistage batch distillation

Major types of batch distillation

Reasons to use batch distillation

1. Small capacity (e.g., specialty chemicals)2. Intermittent need3. Test run for a new product4. 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 distillationno 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- single charge (F) to still pot at time = 0- vapor is withdrawn continuously- composition of liquid in still pot (xW)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, yD, xD

time 0: F, xF

still pot with heater

Rayleigh equation

TMB: F = Wfinal + Dtotal

time t: W, xW y = xD

V, yD, xD

time 0: F, xF

still pot, with heater

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:dCMB: - xDdW = - d(WxW)

(vapor withdrawn) = (change in still pot composition)

chain rule: - xDdW = - WdxW - xWdW

Rayleigh equation where xD = f(xW)

WAIT! K is not constant; K = K(T)

Integration of the Rayleigh equation

Constant relative volatility:

Simpson’s rule:1/

(xD –

xD)

x = xW

Numerical integration:

Specify F, xF, and either Wfinal or xW,final. xW,final

• •

xF

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 distillationAdd 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(vapor withdrawn) = (new solvent added)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

W, xW

V, yD, xD = 1

H2O(l)still

pot,

no h

eate

r

steam

Used for thermally fragile organics (e.g., essential oils in perfume industry), and for slurries/sludges containing organics.

H2O(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.Raoult’s Law: Ptotal = P*WxW +

P*H2O

Both H2O and organic vaporize well below their single-component boiling points.

Fix Ptotal, then T cannot vary!constant T < Tbp(H2O)

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: 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

time 0: F, xF

still pot, with heaterstage N+1

L0, x0

LN, xNVN+1, xN+1

stage 1

y1 ≠ K / xW

stage N

Vj+1, yj+1 Lj, xj

stage j CMO: Vj+1 = Vj and Lj = Lj-1

operating line equation:y = (L/V) x + (1 - (L/V)) xD y = x = xDslope = 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

Choice of operating methods

VLE

y=x

Constant reflux ratio (variable xD)

y=x

Constant distillate composition (variable R)

VLE

•xD•xD

•

•

•

•tim

e time

total reflux

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

VLE

y=x

1. For an arbitrary set of xD 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 xW

3. Perform numerical integration: plot 1/(xD-xW) vs xW

limits: xF, xW,final

4. Calculate Wfinal using Rayleigh equation

• •1

2For N = 2 (incl. reboiler)

Given F, xF, xW,final, R and N, find Dtotal, xD,avg

• xD,4

•xD,1

•xD,2

•xD,3

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.

xW,1

•

•1

2

xW,2

•

•1

2

xW,3

•

•1

2

xW,4

Operating time at constant R (D)

• V = Vmax when vapor velocity u = uflood

• uflood depends on column diameter

• typically, operate at D = 0.75 Dmax

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, 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

VLE

y=x

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)

3. Find xW,min using (L/V) = 1.Rayleigh equation not needed!Verify xW,final > xW,min.

•

N = 2 (incl. reboiler)

•xD

4. Solve mass balance for Wfinal and Dtotal.

Given F, xF, xD (maybe) xw,final and N, find Rinitial, Rmin, xW,min

(L/V)min

N = ∞, R = Rmin

(L/V) initial

xF

•

xW.min

•

Operating time with constant xD

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

mass balance:

x = xW

Numerical integration:

xW,final

• •

xF

Optimal control

• 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

Effect of liquid holdup on the column

• use optimal, time-dependent reflux ratio (not constant R, not constant xD)

• more energy-efficient• useful for difficult separations