### 120712ChE128-4-ColumnDist

```Continuous Column Distillation
Column diagram
total condenser
• to keep the liquid flow rate
constant, part of the distillate
must be returned to the top of
the column as reflux
• the partial reboiler is the last
equilibrium stage in the system
feed
F, xF
feed stage
stripping section
• for a binary mixture, the
compositions xF, xD, xB all
refer to the more volatile
component
reflux
L, xR
L
V
temperature
• all three external streams (F,
D, B) can be liquids (usual case)
reflux drum
(accumulator)
enriching section
• liquid/vapor streams inside
the column flow countercurrent in direct contact with
each other
distillate
D, xD
xR = xD
yB ≠ xB
xD ≠ K xB
boilup
V, yB
partial reboiler
bottoms
B, xB
External mass balance
TMB:
CMB:
F=D+B
F x F = D xD + B xB
for specified F, xF, xD, xB, there
are only 2 unknowns (D, B)
distillate
D, xD
feed
F, xF
æx -x ö
B
D = çç F
÷÷ F
è xD - xB ø
B=F-D
bottoms
B, xB
External energy balance
• assume column is wellinsulated, adiabatic
EB:
distillate
D, xD
F hF + QC + QR
= D hD + B hB
F, hF are known
feed
F, xF
D and B are saturated liquids
so hD, hB are also known
unknowns: QC, QR
• need another equation
bottoms
B, xB
Balance on condenser
1. Mass balance
TMB:
V1 = D + L 0
CMB:
y1 = xD = xR (doesn’t help)
unknowns: V1, L0
specify external reflux ratio R = L0/D
vapor
V1, y1
reflux
L 0 , xR
distillate
D, xD
V1 = D + (L0/D)D = (1 + R)D
2. Energy balance
V1H1 + QC = (D + L0)hD = V1hD
QC = V1(hD – H1)
then calculate QR from column energy balance
hD > H1
QC < 0
QR > 0
Splits
compositions in product streams.
What is the fractional recovery (FR) of
benzene in the distillate?
What is the fractional recovery (FR) of
toluene in the bottoms?
Most volatile component (MVC) is benzene:
xF = 0.46
FRMVC =
xDD 0.99D
=
xF F 0.46F
FRLVC =
(1- xB )B 0.98D
=
(1- xF )F 0.54F
Calculating fractional recoveries
æx -x ö
æ 0.46 - 0.02 ö
0.44
F
B
D = çç
÷÷ F = ç
F
=
(620) = 281
÷
0.97
è 0.99 - 0.02 ø
è xD - xB ø
B = F – D = 620 - 281 = 339
FRMVC
FRLVC
xDD 0.99(281)
=
=
= 0.975
xF F 0.46(620)
(1- xB )B 0.98(339)
=
=
= 0.992
(1- xF )F 0.54(620)
Stage-by-stage analysis
Lewis-Sorel method
Consider the top of the distillation column:
vapor
V1, y1
stage 1
L1
x1
V2
y2
V1, V2 are saturated vapors
L0, L1 are saturated liquids
reflux
L0, x0
distillate
D, xD
Which streams have compositions related by VLE?
V1, L1
They are streams leaving the same equilibrium stage.
K1(T1,P) = y1/x1
How are the compositions of streams V2 and L1 related?
Need to perform balances around stage 1.
Relationships for stage 1
vapor
V1, y1
stage 1
L1
x1
reflux
L0, x0
distillate
D, xD
V2
y2
There are 14 variables:
4 flow rates (L1, V2, L0, V1)
4 compositions (x1, y2, x0, y1)
4 enthalpies (h1, H2, h0, H1)
T1, P
TMB:
CMB:
EB:
VLE:
L0 + V2 = L1 + V1
L0x0 + V2y2 = L1x1 + V1y1
L0h0 + V2H2 = L1h1 + V1H1
K1(T1,P) = y1/x1
We usually specify 10 of them:
P, xD, D, R = L0/D
xD = x0 = y1
V1 = L0 + D
T1 and all 4 enthalpies (by VLE)
4 unknowns (L1, x1, V2, y2) and 4 equations:
problem is completely specified.
Relationships for stage 2
L1,x1
V2,y2
stage 2
L2,x2
V3,y3
TMB: L1 + V3 = L2 + V2
CMB: L1x1 + V3y3 = L2x2 + V2y2
EB:
L1h1 + V3H3 = L2h2 + V2H2
VLE: K2(T2,P) = y2/x2
can solve for 4 unknowns (L2, x2, V3, y3)
and so on… proceed down the column to the reboiler. Very tedious.
Simplifying assumption:
If li (latent heat of vaporization) is not a strong function of composition,
then each mole of vapor condensing on a given stage causes one mole of
liquid to vaporize.
Constant Molal Overflow (CMO): vapor and liquid flow rates are constant
Constant molal overflow
TMB: L1 + V3 = L2 + V2
CMO: V3 - V2 = L2 - L1 = 0
V3 = V2 = V
L2 = L1 = L
We can drop all subscripts on L and V in the upper section of
the column (above the feed stage).
internal reflux ratio: L/V = constant
Rectifying column
Feed enters at the bottom, as a vapor.
No reboiler required.
L, xR
D, xD
Can give very pure distillate; but bottoms
stream will not be very pure.
Mass balance around top of column,
down to and including stage j:
stage j
Vj+1,yj+1
F, xF
CMB: Vj+1yj+1 = Ljxj + DxD
Lj,xj
B, xB
CMO: yj+1 = (L/V) xj + (D/V) xD
D=V-L
yj+1 = (L/V) xj + (1 - L/V) xD
Relates compositions of passing streams.
Lewis analysis of rectifying column
1. Assume CMO (Vj = Vj+1 = V; Lj = Lj-1 = L)
2. Need specified xD; xD = y1
3. Stage 1: use VLE to obtain x1
x1 = y1/K1(T1,P)
4. Use mass balance to obtain y2
y2 = (L/V) x1 + (1 - L/V) xD
5. Stage 2: use VLE to obtain x2
x2 = y2/K2(T2,P)
6. Use mass balance to obtain y3
y3 = (L/V) x2 + (1 - L/V) xD
7. Continue until x = xB
Graphical analysis of rectifying column
equation of the operating line:
y = (L/V) x + (1 - L/V) xD
slope = (L/V)
always positive (compare to flash drum)
•xD = x0
(x0,y1)
plotting the operating line:
yint = (1 - L/V) xD
find a second point on the operating line:
y = x = (L/V) x + (1 - L/V) xD = xD
plot xD on y = x
yint•
recall: xD = xR = x0; the passing stream is y1
• the operating line starts at the point (x0,y1)
• the operating line gives the compositions of all passing streams (xj,yj+1)
McCabe-Thiele analysis: rectifying column
1.
2.
3.
4.
Plot VLE line (yi vs. xi)
Draw the y = x line
Plot xD on y = x
Plot yint = xD (1 – L/V)
L/V  internal reflux ratio, usually not specified
instead, the external reflux ratio (R) is specified
L
L
L
D = R
= D=
(L + D)
V V
R +1
D
D
5. Draw in the operating line
6. Step off stages, alternating between VLE and operating line,
starting at (x0,y1) located at y = x = xD, until you reach x = xB
7. Count the stages.
Ex.: MeOH-H2O rectifying column
NEVER “step” over the VLE line.
Rectifying column with total condenser
Specifications: xD = 0.8, R = 2
Find N required to achieve xB = 0.1
stage 1
(x1,y1)•
stage 2
(x2,y2)•
1. Plot VLE line
2. Draw y=x line
3. Plot xD on y=x
4. Plot yint = xD (1 - L/V)
L/V = R/(R+1) = 2/3
yint = xD(1 - L/V)= 0.8/3 = 0.26
stage 3
(x3,y3)•
yint•
•xD= x0
(x0,y1)
•(x1,y2)
•(x2,y3)
•
5. Draw in operating line
lowest xB possible for this op. line
6. Step off stages from xD to xB
7. Count the stages
N=3
•xB
Limiting cases: rectification
Specifications:
xD = 0.8, vary R = L/D
1. L  0
R = L/D  0 NO REFLUX
L/V  0
Column operates like a single
equilibrium stage.
(Why bother?)
L/V = 0
No reflux!
stage 1
(x1,y1)•
0≤R≤ 
0 ≤ L/V ≤ 1
2. D  0
R = L/D   TOTAL REFLUX
L/V = R/(R+1)  1
(L’Hôpital’s Rule)
Operating line is y=x
Max. distance between VLE and op. line
Max. separation on each equil. stage
Corresponds to Nmin, but no distillate!
L/V = 1
Total reflux!
•xD= x0
(x0,y1)
Minimum reflux ratio
L/V = 0
Specifications:
xD = 0.8, vary R
•
The number of stages N required
to reach the VLE-op. line
intersection point is .
•
•
This represents xB,min for a
particular R.
It also represents Rmin for
this value of xB.
•
•xD= x0
(x0,y1)
Increasing R = L/D
Decreasing D
Decreasing xB (for fixed N)
Rmin for this xB
0 ≤ L/V ≤ 1
•
xB ,min for this R
L/V = 1
0≤R≤ 
Optimum reflux ratio
capital cost
operating (energy) cost
∞ stages
cost/lb
total cost
min. heat
required
Rmin
Ropt
external reflux ratio, R
Rule-of-thumb:
1.05 ≤ Ropt/Rmin ≤ 1.25
Ractual can be specified as a multiple of Rmin
Stripping column
Feed enters at the top, as a liquid.
F, xF
No reflux required.
D, xD
Lk-1,
xk-1
Vk,yk
stage k
Can give very pure bottoms; but distillate
stream will not be very pure.
Mass balance around bottom of column,
up to and including stage k:
CMB: Lk-1xk-1 = Vkyk + BxB
B, xB
CMO: yk = (L/V) xk-1 - (B/V) xB
L=V+B
yk = (L/V) xk-1 + (1 - L/V) xB
Graphical analysis of stripping column
equation of the operating line:
y = (L/V) x + (1 - L/V) xB
slope = L / V
always positive
plotting the operating line:
y = x = (L/V) x + (1 - L/V) xB = xB
plot xB on y = x
(xN+1,yN+1)
PR
•
finding the operating line slope:
L V +B
B
=
= 1+
V
V
V
•xB = xN
(xN+1,yN+2)
(recall V/B is the boilup ratio)
Where is the partial reboiler? Designate this as stage N+1, with xN+1 = xB.
Coordinates of the reboiler: (xN+1,yN+1)
McCabe-Thiele analysis: stripping column
1. Plot VLE line (yi vs. xi)
2. Draw the y = x line
3. Plot xB on y = x
4. Draw in the operating line
5. Step off stages, alternating between VLE and operating line,
starting at (xN+1,yN+2) located at y = x = xB, until you reach x = xD
6. Count the stages.
Ex.: MeOH-H2O stripping column
NEVER step over the VLE line.
Column with partial reboiler
Specifications:
xB = 0.07, V / B = 2
Find N required to achieve xD = 0.55
1. Plot VLE line
2. Draw y=x line
(0.7,1)
•
stage 1
(xN-2,yN-2)
stage 2
••
(xN-1,yN-1) • •(xN-2,yN-2)
stage 3
(xN,yN) •
• (xN-1,yN)
xD,max for this
boilup ratio
3. Plot xB = xN+1 on y = x
4. Draw op. line
L / V = 1+ B / V = 1.5
y = 1 = 1.5x - 0.05 x = 1.05 /1.5 = 0.7
PR•
(xN+1,
yN+1)
• (xN,yN+1)
5. Step off stages starting at PR
6. Stop when you reach x = xD
7. Count the stages.
• xB= xN+1
(xN+1,yN+2)
x
•D
Limiting cases: stripping
Specifications:
xB = 0.07, vary boilup ratio V / B
1.
NO BOILUP
Behaves as if the column
wasn’t even there.
(Why bother?)
1£ L /V £ ¥
¥ ³V / B ³ 0
2. B  0
TOTAL BOILUP
PR•
L /V = 1
TOTAL BOILUP
Operating line is y=x
L /V = ¥
• xB= xN+1
Max. distance between VLE and op. line
Max. separation on each equil. stage
Corresponds to Nmin. But no bottoms product!
NO BOILUP
Minimum boilup ratio
Specifications:
xB = 0.07, vary boilup ratio
yD ,max for
this boilup ratio
•
•
•
The number of stages N required
to reach the VLE-op. line
intersection point is .
This represents yD,max for a
particular boilup ratio.
It also represents the
minimum boilup ratio for
this value of yD.
•
L /V = 1
Total boilup
PR•
L /V = ¥
• xB= xN+1
No boilup
1£ L /V £ ¥
¥ ³V / B ³ 0
McCabe-Thiele analysis
of complete distillation column
Total condenser, partial reboiler
Specifications:
xD = 0.8, R = 2
xB = 0.07, V / B = 2
Find N required
Locate feed stage
Feed enters
on stage 2
stage 2 •
stage 1
•
•xD
•
1. Draw y=x line
2. Plot xD and xB on y=x
3. Draw both op. lines
4. Step off stages starting
at either end, using new op.
line as you cross their
intersection
5. Stop when you reach
the other endpoint
6. Count stages
7. Identify feed stage
PR•
• xB
•
NEVER step over
the VLE line.
Feed condition
• Changing the feed temperature affects internal
flow rates in the column
• If the feed enters as a saturated liquid, the liquid
L
V
F + L +V = L +V
• If the feed enters as a saturated vapor, the vapor
flow rate above the feed stage will increase:
V =V + F
• If the feed flashes as it enters the feed stage to form a
two-phase mixture, 50 % liquid, both the liquid and
vapor flow rates will increase:
L = L + 0.5F and
V
feed
F
flow rate below the feed stage will increase:
L = L+F
L
V = V + 0.5F
Feed quality, q
EB:
FhF + LhL +VHV = LhL +VHV
rearrange:
FhF + (V -V)HV = (L - L)hL
TMB:
V -V = L - L - F
substitute:
FhF + (L - L - F)HV = (L - L)hL
combine terms:
(L - L)(HV - hL ) = F(HV - hF )
define:
L - L HV - hF º q q  mol sat’d liquid
=
generated on feed
F
HV - hL
plate, per mol feed
Different types of feed quality
L -L
qº
F
L = L + qF and V = V + (1- q)F
saturated liquid feed
L = L+F
q=1
saturated vapor feed
V =V + F
q=0
feed flashes to form 2-phase
mixture, q% liquid
L = L + qF
0<q<1
subcooled liquid feed
q>1
- some vapor condenses on feed plate
superheated vapor
- some liquid vaporizes on feed plate
q<0
Equation of the feed line
rectifying section CMB:
Vy j+1 = Lx j + DxD
stripping section CMB:
Vy k = Lxk-1 - BxB
intersection of top and
bottom operating lines:
(V -V)y = (L - L)x - (BxB + DxD )
substitute:
BxB + DxD = FzF
æ L -L ö
FzF
y = -ç
÷x +
èV -V ø V -V
L - L = qF and
equation of the feed line:
zF
q
y =x+
(1- q)
1- q
V -V = (1- q)F
Plotting the feed line
sat'd liq
zF
q
y =x+
(1- q)
1- q
sat'd vapor
•zF
where does the feed line intersect y=x?
zF
q
x =x+
(1- q)
1- q
æ
q ö zF
x ç1+
÷=
è (1- q) ø 1- q
zF
x
=
y = x = zF
1- q 1- q
feed type
sat'd liquid
sat'd vapor
2-phase liq/vap
subcooled liq
superheated vap
q slope, m
q=1
m=
q=0
m=0
0<q<1 m < 0
q>1
m>1
q<0
0<m<1
Ex.: Complete MeOH-H2O column
Total condenser, partial reboiler
Specifications:
xD = 0.9, xB = 0.04, zF = 0.5, R=1
Feed is a 2-phase mixture, 50% liq.
Find N and NF,opt.
N = 6 + PR
4
•
3
•
5•
1. Draw y=x line
•zF
2. Plot xD, xB and zF on y=x
6
•
3. Draw feed line, slope = -0.5
2•
1
•
•xD
Operating
lines intersect
on stage 4.
This is NF,opt.
4. Draw top op. line, slope = L/V = 0.5
5. Draw bottom op. line (no
calc. required)
6. Step off stages starting at
either end, using new op. line as
you cross the feed line.
PR •
• xB
We can independently specify
only 2 of the following 3
variables: R, q, V/B (usually: R, q).
Using a non-optimal feed location reduces separation.
Feed lines in rectifying/stripping columns
stripping column
rectifying column
•
•
•
•xD
•
•
•
• yD
•z
F
•
•z
•F
•xB
•
• •
bottom operating line
total condenser, no reboiler
sat’d vapor feed, liquid bottoms
F and B are passing streams
PR•
•
top operating
line
• xB
partial reboiler, no condenser
sat’d liquid feed, vapor distillate
F and V are passing streams
Design freedom
Fixed q. Vary R:
Fixed R. Vary q:
Rmin
•xD
•xD
pinch
point
decrease R
•zF
• xB
choice of R dictates
required boilup ratio.
•zF
qmin
pinch point
• xB
You cannot “step” over a pinch point – this would require N = . It corresponds to a position in
the column where there is no difference in composition between adjacent stages.
Another type of pinch point
Ethanol-water
xD = 0.82, xB = 0.07
zF = 0.5, q = 0.5
Find Rmin
pinch
point
1. Draw y=x line
2. Plot xD, xB and zF on y=x
•zF
3. Draw feed line, slope = q/(q-1)
4. Draw top op. line to intersect
with feed line on VLE line
5. Don’t cross the VLE line!
6. Redraw top operating line as
tangent to VLE.
• xB
•xD
Column with
two feeds:
Column with
three products:
L
V
L´
V´
feed 2
F2, z2, q2
feed 1
F1, z1, q1
L
distillate
D, xD
z2 > z1
and/or
q2 > q 1
V
L
V
L´
V´
feed
F, z
L
bottoms
B, xB
V
distillate
D, xD
side-stream
S, xS or yS
side-streams
must be
saturated
liquid or vapor
bottoms
B, xB
Each intermediate input/output stream changes the mass balance, requiring a new operating line.
Multiple feedstreams
Total condenser, partial reboiler
Specifications:
xD = 0.9, xB = 0.07, z1 = 0.4, z2=0.6
Some specified q-values
R = 1. Find N, NF1,opt, NF2,opt
3•
2. Plot xD, z1, z2 and xB on y=x
•z2
5•
•z1 = z2
3. Draw both feed lines
•z1
4. Draw top op. line, slope = L/V
7. Step off stages starting
at either end, using new op. line each
time you cross an intersection
•xD
4
•
1. Draw y=x line
5. Calculate slope of middle
operating line, L´/V´, and draw
middle operating line
6. Draw bottom operating line
(no calc. required)
2•
1
•
PR
•
x•B
Optimum location
for feed 1 is stage 5.
Optimum location
for feed 2 is stage 3.
Slope of middle operating line
2-feed mass balances:
TMB: F2 + V´ = L´ + D
CMB: F2z2 + V´yj+1 = L´xj + DxD
middle operating line equation:
y = (L´/V´)x + (DxD - F2z2)/V´
obtain slope from:
L´ = F2q2 + L = F2q2 + (R)(D)
V´ = L´ + D – F2
feed 2
F2, z2, q2
D, xD
stage j
L´
V´
side-stream  feed-stream with –ve flow rate
sat’d liq
y = x = xS
sat’d vapor y = x = yS
D, xD
side-stream mass balances:
TMB: V´- L´= D + S
CMB: V´yj+1 - L´xj = DxD + SxS
stage j
middle operating line equation:
y = (L´/V´)x + (DxD + SxS)/V´
L´
side-stream
S, xS or yS
V´
McCabe-Thiele analysis of side-streams
Saturated liquid side-stream, xs = 0.64
•
•
Saturated vapor side-stream, ys = 0.73
•xD
•
•
•
•yS
•xS
•z
•x
B
•z
•x
B
Side-stream must correspond exactly to stage position.
•xD
Partial condensers
A partial condenser can be used when a
vapor distillate is desired:
2
•
D, yD
V, y1
L, x0
L
V
A partial condenser is an equilibrium stage.
CMB: Vyj+1 = Lxj + DyD
Operating line equation:
y = (L/V)x + DyD = (L/V)x + (1 - L/V)yD
1
•
PC
•
•yD
Total reboilers
A total reboiler is simpler (less expensive)
than a partial reboiler and is used when the
N-1
•
V
L
stage N
N•
V, yB
B, xB
A total reboiler is not an equilibrium
stage.
TR •xB,yB
Stage efficiency
Under real operating conditions, equilibrium is approached but not achieved:
Nactual > Nequil
overall column efficiency:
Eoverall = Nequil/Nactual
Efficiency can vary from stage to stage.
Reboiler efficiency ≠ tray efficiency
Murphree vapor efficiency:
EMV =
y n - y n+1
y n * -y n+1
where yn* is the equilibrium vapor composition
(not actually achieved) on stage n:
yn* = Kn xn
Can also define Murphree liquid
efficiency:
EML =
x n - x n-1
x n * -x n-1
xj* = yj / Kj
Ex.: Vapor efficiency of MeOH-H2O column
Total condenser, partial reboiler
Specifications:
xD = 0.9, xB = 0.07, z = 0.5, q = 0.5,
R = 1, EMV,PR = 1, EMV = 0.75.
Find N and NF,opt.
•
5
•

6
•
1. Draw y=x line
•
4
•
3
•
2
•
1 •xD
7
2. Plot xD, z, and xB on y=x
•
3. Draw feed line
•z

8
4. Draw top op. line, slope = L/V
5. Draw bottom operating line
(no calc. required)
PR•
NF,opt = 6
6. Find partial reboiler
7. Step off stages, using EMV to
N = 8 + PR
• xB
8. Label real stages.
Intermediate condensers and reboilers
Intermediate condensers/reboilers can improve
the energy efficiency of column distillation:
1. by decreasing the heat that must be supplied
at the bottom of the column, providing part of
the heat using an intermediate reboiler
- use a smaller (cheaper) heating element at the
bottom of the column, or lower temperature
steam to heat the boilup
1. by decreasing the cooling that must be
supplied at the top of the column, providing
part of the cooling using an intermediate
2.
- use a smaller (cheaper) cooling element at the top
of the column, and/or a higher temperature coolant
for the intermediate condenser
Each column section has its own operating line.
L
V
L´
V´
distillate
D, xD
feed
F, z
S, xS
L´´
intermediate
reboiler
V´´
yS = xS
L
V
bottoms
B, xB
Subcooled reflux
If the condenser is located below the top of the
column, the reflux stream has to be pumped to
the top of the column.
Pumping a saturated liquid damages the
pump, by causing cavitation. The reflux
stream (L0) should be subcooled. This will
cause some vapor to condense.
V1 = V2 - c
and
stage 1
c
V2
L0, x0
L1
L1 = L0 + c
D, xD
CMO is valid below stage 1. Find L/V = L1/V2?
c=
V1, y1
h - h0
L = (1- q0 )L0 q  quality of reflux
0
H -h 0
(
)
2 - q0 L0 / V1
L1 L0 + c L0 + (1- q0 )L0
=
=
=
V2 V1 + c V1 + (1- q0 )L0 1+ (1- q0 )L0 / V1
where L0/V1 = (L0/D)/(1 + L0/D) = R/(R + 1)
EB: V2H2 + L0h0 = V1H1 + L1h1
where H1  H2 = H, but h0 ≠ h1 = h
(V2 – V1)H = L1h - L0h0
cH = (L0 + c)h - L0h0 = L0(h - h0) + ch
Subcooled reflux causes L/V to increase.
Superheated boilup causes L/V to increase.
Open steam distillation
If the bottoms stream is primarily water,
then the boilup is primarily steam.
Can replace reboiler with direct steam
heating (S).
L, xR D, xD
mostly MeOH
MeOH/H2O
feed
F, z
Top operating line and feed lines do not
change.
Bottom operating line is different:
stage j
TMB:
Vj+1,
yj+1
S, yS
Lj,
xj
B, xB
V + B = L + S
usually 0
mostly H2O
bottoms
B, xB
CMB:
V yj+1 + B xB = L xj + S yS
CMO:
B=L
Operating line equation:
y = (L/V) x - (L/V) xB
xint: x = xB
Ex.: Open steam distillation of MeOH/H2O
Specifications:
xD = 0.9, xB = 0.07, zF = 0.5
Feed is a 2-phase mixture, 50% liq.
Total condenser, open steam, R = 1.
Find N and NF,opt.
3
•
2
•
1
•
•xD
4•
1. Draw y=x line
5•
2. Plot xD and zF on y=x
•zF
3. Plot xB on x-axis
4. Draw feed line, slope = q/(q-1)
5. Draw top op. line, slope = L/V
6. Draw bottom op. line (no
calc. required)
7. Step off stages starting
at either end, using new op. line
as you cross their intersection
6
•
All stages are on the column
(no partial reboiler).
N=6
x•B
NF,opt = 4
Column internals
Sieve tray
•
•
•
•
Also called a perforated tray
Simple, cheap, easy to clean
Good for feeds that contain suspended solids
Poor turndown performance (low efficiency when operated below designed flow rate);
prone to “weeping”
Other types of trays
Valve tray
• Some valves close when vapor velocity drops,
keeping vapor flow rate constant
• Better turndown performance
• Slightly more expensive, and harder to clean
than sieve tray
Bubble cap tray
• Excellent contact between vapor and liquid
• Risers around holes prevent weeping
• Good performance at high and low liquid
flow rates
• Very expensive, and very hard to clean
• Not much used anymore
Downcomers
Dual-flow tray (no downcomer)
• Both liquid and vapor pass through holes
• Narrow operating range
In large diameter columns, use multi-pass trays
Cross-flow tray (single pass)
Dual-pass tray
Tray efficiency
efficiency
design
point

flooding
weeping/du
mping
vapor flow rate
• Weeping/dumping: when vapor flow rate is too low, liquid drips
constantly/periodically through holes in sieve tray
• Flooding: when vapor flow rate is too high, liquid on tray mixes with liquid on
tray above
Column distillation videos
Normal column operation:
Column flooding
1. jet flood (due to entrainment)
3. insufficient downcomer clearance
2. lack of downcomer seal
• vapor flow rate too high
• weir height below downcomer
• vapor flows up downcomer
• weir height above downcomer
• liquid backs up downcomer
• ensure bottom edge of downcomer is 1⁄2´´ below top edge of outlet weir.
Column sizing
1. Calculate vapor flood velocity, uflood (ft/s)
æs ö
= Csb,f ç ÷
è 20 ø
0.2
uflood
rL - rV
rV
where Csb,f is the capacity factor, from empirical correlation with flow parameter, FP
WL
FP =
WV
rV
rL
where WL and WV are the mass flow rates of liquid and vapor, respectively
2. Determine net area required for vapor flow, Anet, based on
operating vapor velocity, uop, ft/s
V (MWV )
uop = 0.75uflood =
3600rV Anet
where V is molar vapor flow rate and MWV is average molecular weight of vapor
Tray spacing
Column sizing, cont.
Relationship between net area for vapor flow, Anet, in ft2, and
column diameter, D, in ft:
p D 2h
Anet =
4
where h is the fraction of the cross-sectional area available for vapor flow (i.e., not
occupied by the downcomer)
The required column diameter, D, in ft, is also:
D=
(
4V MWV
)
3600phrV uop
4V
RT
»
3600phuop P
Required column diameter changes where the mass balance changes.
- build column in sections, with optimum diameter for each section, or
- build column with single diameter:
if feed is saturated liquid, design for the bottom
if feed is saturated vapor, design for the top
- balance section diameters (2-enthalpy feed, intermediate condenser/reboiler)
Packed columns
structured packing:
random packing:
• larger surface area, for better contact between liquid and vapor
• preferred for column diameters < 2.5´
• packing is considerably more expensive than trays
• change in vapor/liquid composition is continuous (unlike staged column)
• analysis like a staged column: HETP (= Height Equivalent to a Theoretical Plate/Tray)
packing height required = no. equil. stages x HETP
• packing can be metal, ceramic, glass (depends on feed requirements: corrosive, high T, etc)
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