Batch Distillation
History
Early distillation of alcohol
http://essentialspirits.com/history.htm
Alembic still for distillation of brandy
Major types of batch distillation
Simple batch distillation
Multistage batch distillation
Reasons to use batch distillation
1.
2.
3.
4.
5.
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)
6. Feed varies widely in composition
Simple batch distillation
no rectification ( = no column)
Characteristics:
V, y
D, xD
time 0: F, xF
time t: W, xW
still pot with heater
- 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- vapor is withdrawn continuously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
Rayleigh equation
TMB: F = Wfinal + Dtotal
CMB: FxF = WfinalxW,final + DtotalxD,avg
V, y
D, xD
time 0: F, xF
time t: W, xW
y = xD
still pot, with heater
y
x
K= D = D
xW xW
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:
(vapor withdrawn) = (change in still pot
composition)
WAIT! K is not constant;
K = K(T)
Rayleigh equation
- xDdW = - d(WxW)
chain rule:
ò
Wfinal
F
- xDdW = - WdxW - xWdW
xW ,final
dxW
dW
=ò
xF
W
xD - xW
where xD = f(xW)
Integration of the Rayleigh equation
ò
Wfinal
F
xW ,final
xF
dxW
dxW
dW
=ò
= -ò
xF
xW ,final x - x
W
xD - xW
D
W
Numerical integration:
area =
Constant relative volatility:
a xW
1+ (a -1)xW
(
éx
1- xF
Wfinal
1
W ,final
ê
ln
=
ln
F
a -1 êë xF 1- xW ,final
(
) ùú + lnéê 1- x
) úû êë1- x
1/(xD – xD)
y D = xD =
ù
ú
û
W ,final ú
Fx F -Wfinal xW ,final
F -Wfinal
Dtotal = F -Wfinal
xW ,final
•
•
xW,final
x = xW
xF
dxW
xD - xW
F
Specify F, xF, and either Wfinal or xW,final.
xD,avg =
ò
xF
Simpson’s rule:
area =
ò
x2
x1
ù
æx +x ö
x 2 - x1 é
1
2
êf x1 + 4f ç
f (x) =
÷ + f x2 ú
6 êë
úû
è 2 ø
( )
Wfinal = Fe-Area
( )
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 = - xDdS = WdxW
(W constant)
ò
S
0
x
dxW
dS
=-ò F
xW ,final x
W
D
S 1 é xF ù a -1
ú+
= lnê
xF - xW ,final
W a êë xW ,final úû a
(
)
Batch steam distillation
Used for thermally fragile organics (e.g., essential
oils in perfume industry), and for slurries/sludges
containing organics.
If W, D are immiscible with water, we have
a heterogeneous azeotrope.
D.o.F. = 2 components – 3 phases + 2 = 1
Fix Ptotal, then T cannot vary!
constant T < Tbp(H2O)
Both H2O and organic vaporize well below their
single-component boiling points.
Also, constant vapor composition.
Raoult’s Law:
Ptotal = P*WxW + P*H2O
V, y
still pot,
no heater
A single charge (F) added to still pot at
time = 0. Steam is added continuously.
D, xD = 1
W, xW
H2O(l)
H2O(l)
(to waste)
steam
How much steam is required?
*
nH 2O y H 2O PH 2O Ptotal - Porg xW
=
=
=
*
norg
y org
Porg
Porg
xW
Steam also needed to heat and
vaporize the material in the still pot.
Batch distillation with rectification
V1, y1
stage 1
Lj, xj
stage N
VN+1, xN+1
D, xD
y1 = x 0 = xD
y1 ≠ K / xW
stage j
Vj+1, yj+1
L0, x0
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
LN, xN
time 0: F, xF
time t: W, xW
still pot, with heater
stage N+1
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
Choice of operating methods
Constant reflux ratio (variable xD)
Constant distillate composition (variable R)
•xD
•xD
•
•
•
•
distillation must end
when (or before) xD,avg = xF
Easy to monitor and control.
distillation must end
when (or before) R = ∞ (L/V = 1)
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
For N = 2 (incl.
reboiler)
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
•
2
1 •
•
2
•
1•
2•
•xD,2
•xD,3
•xD,4
2•
xW,4 xW,3
xW,2
xW,1
5. Solve mass balances for
Dtotal and xD,avg
If xD,avg is specified instead of xW,final:
1
•
guess xW,final, calculate xD,avg, iterate.
•xD,1
Operating time at constant R (D)
tbatch = toperating +tdown
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
toperating =
Dtotal F -Wfinal
F -Wfinal
=
=
D
V -L
V(1- L / V)
1
1- L /V =
R +1
toperating =
condenser TMB:
D=
V
R +1
Dmax =
Vmax
R +1
(R +1)
F -Wfinal
V
(
)
• 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.
uflood
æ s ö0.2 r - r
L
V
= Csb,flood ç ÷
rV
è 20 ø
where σ is surface tension,  L 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.
diameter (feet) =
(
4V MWv
)
phrVuop (3600)
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
N = ∞, R = Rmin
•
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)
•xD
•
•
3. Find xW,min using (L/V) = 1.
Rayleigh equation not needed!
Verify xW,final > xW,min.
4. Solve mass balance for Wfinal
and Dtotal.
xW.min
xF
N = 2 (incl. reboiler)
Operating time with constant xD
mass balance:
xD - xF
W =F
xD - xW
Numerical integration:
area =
dW
xD - xF dxw
=F
2 dt
dt
xD - xw
(
)
ò
dxW
xF
xW ,final
(
(1- L / V ) xD - xw
1
(
(1- L / V ) xD - xw
)
2
•
•
xW,final
x = xW
xF
dW dD
=
= V - L = V(1- L /V)
dt
dt
toperating
x - xF
=F D
V
ò
xF
xW ,final
dxW
(
(1- L / V) xD - xw
)
2
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
)
2
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