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Continuous Column Distillation
Column diagram
feedF, xF
bottomsB, xB
boilupV, yB
refluxL, xR
distillateD, xD
feed stage
total condenser
reflux drum(accumulator)
VL
• liquid/vapor streams inside the column flow counter-current in direct contact with each other
partial reboiler
xR = xD
yB ≠ xB
xD ≠ K xBtem
pera
ture
• all three external streams (F, D, B) can be liquids (usual case)
• for a binary mixture, the compositions xF, xD, xB all refer to the more volatile component
• 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
enric
hing
sec
tion
strip
ping
sec
tion
External mass balance
TMB: F = D + BCMB: F xF = D xD + B xB
for specified F, xF, xD, xB, there are only 2 unknowns (D, B)
B = F - D
feedF, xF
bottomsB, xB
distillateD, xD
External energy balance• assume column is well-
insulated, adiabatic
feedF, xF
bottomsB, xB
distillateD, xD
EB: F hF + QC + QR
= D hD + B hBF, hF are known
D and B are saturated liquidsso hD, hB are also known
unknowns: QC, QR
• need another equation
Balance on condenser
distillateD, xD
refluxL0, xR
vaporV1, y1
1. Mass balanceTMB: V1 = D + L0
CMB: y1 = xD = xR (doesn’t help)
unknowns: V1, L0
specify external reflux ratio R = L0/D
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 < 0QR > 0
SplitsSometimes used instead of specifying 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
Calculating fractional recoveries
B = F – D = 620 - 281 = 339
vaporV1, y1
distillateD, xD
Stage-by-stage analysisLewis-Sorel method
refluxL0, x0
V2
y2
L1
x1
stage 1
Consider the top of the distillation column:
V1, V2 are saturated vaporsL0, L1 are saturated liquids
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
TMB: L0 + V2 = L1 + V1
CMB: L0x0 + V2y2 = L1x1 + V1y1
EB: L0h0 + V2H2 = L1h1 + V1H1
VLE: K1(T1,P) = y1/x1
vaporV1, y1
distillateD, xD
refluxL0, x0
V2
y2
L1
x1
stage 1
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
We usually specify 10 of them:P, xD, D, R = L0/DxD = x0 = y1
V1 = L0 + DT1 and all 4 enthalpies (by VLE)
4 unknowns (L1, x1, V2, y2) and 4 equations:problem is completely specified.
Relationships for stage 2
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)
V2,y2L1,x1
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
stage 2
V3,y3L2,x2
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
B, xB
D, xD
F, xF
Feed enters at the bottom, as a vapor.
No reboiler required.
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:
L, xR
stage j
Vj+1,yj+1 Lj,xj
CMB: Vj+1yj+1 = Ljxj + DxD
CMO: yj+1 = (L/V) xj + (D/V) xD
D = V - Lyj+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 columnequation of the operating line:y = (L/V) x + (1 - L/V) xD
slope = (L/V)always positive (compare to flash drum)
plotting the operating line:yint = (1 - L/V) xD
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)
find a second point on the operating line:y = x = (L/V) x + (1 - L/V) xD = xD
plot xD on y = x
VLE
y=x
yint•
op. line
•xD = x0
(x0,y1)
McCabe-Thiele analysis: rectifying column1. Plot VLE line (yi vs. xi)
2. Draw the y = x line3. Plot xD on y = x
4. Plot yint = xD (1 – L/V)L/V internal reflux ratio, usually not specifiedinstead, the external reflux ratio (R) is specified
5. Draw in the operating line6. 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
VLE
y=x
yint•
op. line
(x1,y1)•
(x2,y2)•
(x3,y3)•
•(x1,y2)
•(x2,y3)
•xB
2. Draw y=x line
3. Plot xD on y=x
4. Plot yint = xD (1 - L/V)
6. Step off stages from xD to xB
7. Count the stages
Rectifying column with total condenserSpecifications: xD = 0.8, R = 2Find N required to achieve xB = 0.1
L/V = R/(R+1) = 2/3yint = xD(1 - L/V)= 0.8/3 = 0.26
NEVER “step” over the VLE line.
lowest xB possible for this op. line
•
1. Plot VLE line
5. Draw in operating line
•xD= x0
(x0,y1)
stage 1
stage 2
stage 3
N = 3
Limiting cases: rectification
•xD= x0
(x0,y1)
VLE
y=x
stage 1(x1,y1)•
Specifications:xD = 0.8, vary R = L/D
1. L 0 R = L/D 0 NO REFLUXL/V 0
L/V = 0No reflux!
Max. distance between VLE and op. lineMax. separation on each equil. stageCorresponds to Nmin, but no distillate!
L/V = 1Total reflux!
2. D 0 R = L/D TOTAL REFLUXL/V = R/(R+1) 1(L’Hôpital’s Rule)
Operating line is y=x
Column operates like a single equilibrium stage. (Why bother?)
0 ≤ L/V ≤ 10 ≤ R ≤
Minimum reflux ratio
VLE
y=x
•
Specifications:xD = 0.8, vary R
L/V = 0
The number of stages N required to reach the VLE-op. line intersection point is .
L/V = 1
0 ≤ L/V ≤ 1
0 ≤ R ≤
•
•
•
•xB ,min for this R
This represents xB,min for a particular R.
It also represents Rmin for this value of xB.
•xD= x0
(x0,y1)
Rmin for this xB
Increasing R = L/DDecreasing DDecreasing xB (for fixed N)
Optimum reflux ratio
cost
/lb
external reflux ratio, R
Rmin Ropt
operating (energy) cost
Rule-of-thumb:1.05 ≤ Ropt/Rmin ≤ 1.25
Ractual can be specified as a multiple of Rmin
∞ st
ages
min. heat required
capital cost
total cost
Stripping column
B, xB
D, xD
F, xF
Feed enters at the top, as a liquid.
No reflux required.
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:
stage k
CMB: Lk-1xk-1 = Vkyk + BxB
Vk,ykLk-1,xk-1
CMO: yk = (L/V) xk-1 - (B/V) xB
L = V + Byk = (L/V) xk-1 + (1 - L/V) xB
Graphical analysis of stripping column
Where is the partial reboiler? Designate this as stage N+1, with xN+1 = xB.
VLE
y=xop
. lin
e
•xB = xN
(xN+1,yN+2)
equation of the operating line:y = (L/V) x + (1 - L/V) xB
slope = L / Valways positive
plotting the operating line:y = x = (L/V) x + (1 - L/V) xB = xB
plot xB on y = x
finding the operating line slope:
(recall V/B is the boilup ratio)
Coordinates of the reboiler: (xN+1,yN+1)
(xN+1,yN+1)PR•
McCabe-Thiele analysis: stripping column1. 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
VLE
y=x
(0.7,1) •
op. li
ne
•(xN-2,yN-2)stage 2
2. Draw y=x line
3. Plot xB = xN+1 on y = x
4. Draw op. line
5. Step off stages starting at PR
6. Stop when you reach x = xD
Column with partial reboilerSpecifications:xB = 0.07, Find N required to achieve xD = 0.55
7. Count the stages.
(xN+1,yN+1)
• (xN,yN+1)•PR
(xN,yN) • • (xN-1,yN)
(xN-1,yN-1) •
xD,max for this boilup ratio
•
stage 3
NEVER step over the VLE line.
(xN-2,yN-2) •
stage 1
• xB= xN+1(xN+1,yN+2)
1. Plot VLE line
•xD
Limiting cases: stripping
VLE
y=x
Specifications:xB = 0.07, vary boilup ratio
Max. distance between VLE and op. lineMax. separation on each equil. stageCorresponds to Nmin. But no bottoms product!
Behaves as if the column wasn’t even there. (Why bother?)
• xB= xN+1
•PR
1. NO BOILUP
2. B 0
Operating line is y=x
TOTAL BOILUP
NO BOILUP
TOTAL BOILUP
Minimum boilup ratio
VLE
y=x
Specifications:xB = 0.07, vary boilup ratio
The number of stages N required to reach the VLE-op. line intersection point is .
Increasing boilup ratio
Decreasing B
Increasing x D (fo
r fixed N)
yD ,max for this boilup ratio•
This represents yD,max for a particular boilup ratio.
It also represents the minimum boilup ratio for this value of yD.
• xB= xN+1
•PR
No boilup
Total boilup
•
•
•
McCabe-Thiele analysis of complete distillation column
VLE
y=x
stage 1
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
Total condenser, partial reboilerSpecifications:xD = 0.8, R = 2xB = 0.07, Find N requiredLocate feed stage
6. Count stages
••PR
• •
•
stage 2
• xB
•xD
top o
p. lin
e
bottom op. li
ne
Feed enters on stage 2
NEVER step over the VLE line.
7. Identify feed stage
• If the feed enters as a saturated liquid, the liquid
flow rate below the feed stage will increase:
• If the feed enters as a saturated vapor, the vapor
flow rate above the feed stage will increase:
Feed condition• Changing the feed temperature affects internal
flow rates in the columnVL
feedF
VL
and
• 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:
Feed quality, q
EB:
rearrange:
TMB:
substitute:
combine terms:
define:q mol sat’d liquid generated on feed plate, per mol feed
Different types of feed quality
saturated liquid feed q = 1
feed flashes to form 2-phase0 < q < 1
mixture, q% liquid
and
subcooled liquid feedq > 1
- some vapor condenses on feed plate
superheated vapor q < 0
- some liquid vaporizes on feed plate
saturated vapor feedq = 0
Equation of the feed line
rectifying section CMB:
and
stripping section CMB:
intersection of top and bottom operating lines:
substitute:
equation of the feed line:
Plotting the feed line
where does the feed line intersect y=x?
y=x
VLE
y = x = zF
feed type q slope, msat'd liquid q=1 m = sat'd vapor q=0 m = 02-phase liq/vap 0<q<1 m < 0subcooled liq q>1 m > 1superheated vap q<0 0<m<1
•zF
sat'd
liq
sat'd vapor
2-phase
subc
oole
d liq
superheated vap
Ex.: Complete MeOH-H2O column
y=x
1. Draw y=x line
2. Plot xD, xB and zF on y=x
3. Draw feed line, slope = -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.
Total condenser, partial reboilerSpecifications:xD = 0.9, xB = 0.04, zF = 0.5, R=1Feed is a 2-phase mixture, 50% liq.Find N and NF,opt.
•
4. Draw top op. line, slope = L/V = 0.5
• xB
•xD
•zF
• •
•
•
•1
6
2
5
34
Operating lines intersect on stage 4. This is NF,opt.
Using a non-optimal feed location reduces separation.
N = 6 + PR
•
PRWe can independently specify only 2 of the following 3 variables: R, q, V/B (usually: R, q).
Feed lines in rectifying/stripping columns
•
••PR
• •
•
• xB
•zF
•
• •
•
•zF
•xD
stripping column
partial reboiler, no condensersat’d liquid feed, vapor distillateF and V are passing streams
rectifying column
total condenser, no reboilersat’d vapor feed, liquid bottomsF and B are passing streams
•
•xB
•
bottom operating line
top operating line
•
• yD
Design freedom
choice of R dictates required boilup ratio.
Fixed q. Vary R:
• xB
•xD
•zF
Fixed R. Vary q:
• xB
•xD
•zF
Rmin
heat t
he fee
d
qmin
pinch point
decrease R
pinch point
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
VLE
y=x
1. Draw y=x line
2. Plot xD, xB and zF on y=x
3. Draw feed line, slope = q/(q-1)
5. Don’t cross the VLE line!
Ethanol-water xD = 0.82, xB = 0.07zF = 0.5, q = 0.5Find Rmin
4. Draw top op. line to intersect with feed line on VLE line
• xB
•xD
•zF
6. Redraw top operating line as tangent to VLE.
pinch point
Additional column inputs/outputs
bottomsB, xB
distillateD, xD
VL
feed 2F2, z2, q2
V´L´
feed 1F1, z1, q1
VL
distillateD, xD
bottomsB, xB
feedF, z
VL
side-streamS, xS or yS
V´L´
VL
Column with two feeds:
Column with three products:
Each intermediate input/output stream changes the mass balance, requiring a new operating line.
z2 > z1
and/or q2 > q1
side-streams must be saturated liquid or vapor
Multiple feedstreams
VLE
y=x
1. Draw y=x line
2. Plot xD, z1, z2 and xB on y=x
3. Draw both feed lines
6. Draw bottom operating line (no calc. required) 7. Step off stages starting at either end, using new op. line each time you cross an intersection
4. Draw top op. line, slope = L/V
•
• •
•
•
•1
PR
2
5
34
Optimum location for feed 1 is stage 5.Optimum location for feed 2 is stage 3.
Total condenser, partial reboilerSpecifications:xD = 0.9, xB = 0.07, z1 = 0.4, z2=0.6Some specified q-valuesR = 1. Find N, NF1,opt, NF2,opt
• xB
•xD
•z2
•z1
5. Calculate slope of middle operating line, L´/V´, and draw middle operating line
•z1 = z2
Slope of middle operating line2-feed mass balances:TMB: F2 + V´ = L´ + DCMB: 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 2F2, z2, q2
D, xD
V´L´
stage j
side-streamS, xS or yS
D, xD
V´L´
stage j
middle operating line equation: y = (L´/V´)x + (DxD + SxS)/V´
side-stream feed-stream with –ve flow rate sat’d liq y = x = xS
sat’d vapor y = x = yS
side-stream mass balances:TMB: V´- L´= D + SCMB: V´yj+1 - L´xj = DxD + SxS
McCabe-Thiele analysis of side-streams
VLE
y=x
Saturated liquid side-stream, xs = 0.64
• xB
•xD
•xS
•z
Saturated vapor side-stream, ys = 0.73
VLE
y=x
• xB
•xD
•yS
•z
Side-stream must correspond exactly to stage position.
• •
• •
•
Partial condensers
A partial condenser can be used when a vapor distillate is desired:
D, yD
VL
L, x0
V, y1
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
y=x
•yD
•
PC1
2 •
•
Total reboilersA total reboiler is simpler (less expensive) than a partial reboiler and is used when the bottoms stream is readily vaporized:
A total reboiler is not an equilibrium stage.
y=x
•xB,yBTR
N
N-1 •
•
B, xB
V L
stage N
V, yB
Stage efficiencyUnder 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
where yn* is the equilibrium vapor composition (not actually achieved) on stage n:
yn* = Kn xn
Can also define Murphree liquid efficiency:
xj* = yj / Kj
Murphree vapor efficiency:
y=x
• xB
•xD
•z
•
•
PR
•
Ex.: Vapor efficiency of MeOH-H2O columnTotal condenser, partial reboilerSpecifications: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.
•
• •
•
1. Draw y=x line
2. Plot xD, z, and xB on y=x
3. Draw feed line
5. Draw bottom operating line (no calc. required)
6. Find partial reboiler
4. Draw top op. line, slope = L/V
7. Step off stages, using EMV to adjust vertical step size.8. Label real stages.
•
2
8
3
6
45
1
7•
N = 8 + PR
NF,opt = 6
To use ELV, adjust horizontal step size instead.
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 instead
- use a smaller (cheaper) heating element at the bottom of the column, or lower temperature steam to heat the boilup
2. by decreasing the cooling that must be supplied at the top of the column, providing part of the cooling using an intermediate condenser instead
3. - use a smaller (cheaper) cooling element at the top of the column, and/or a higher temperature coolant for the intermediate condenser
distillateD, xD
bottomsB, xB
feedF, z
VL
VL
V´L´S, xS
intermediate reboiler
yS = xS
Each column section has its own operating line.
V´´L´´
V2 L1
L0, x0
V1, y1
D, xD
stage 1
Subcooled refluxIf the condenser is located below the top of the column, the reflux stream has to be pumped to the top of the column.
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
where L0/V1 = (L0/D)/(1 + L0/D) = R/(R + 1)
CMO is valid below stage 1. Find L/V = L1/V2?
V1 = V2 - c and L1 = L0 + c
cPumping a saturated liquid damages the pump, by causing cavitation. The reflux stream (L0) should be subcooled. This will cause some vapor to condense.
Subcooled reflux causes L/V to increase.Superheated boilup causes L/V to increase.
q0 quality of reflux
Open steam distillation
B, xB
D, xD
mostly MeOH
S, yS
L, xR
stage j
Vj+1,yj+1
Lj,xj
MeOH/H2OfeedF, z
bottomsB, xB
If the bottoms stream is primarily water, then the boilup is primarily steam.
Can replace reboiler with direct steam heating (S).
Top operating line and feed lines do not change.
Bottom operating line is different:
TMB: V + B = L + S
CMB: V yj+1 + B xB = L xj + S yS
usually 0
Operating line equation:
y = (L/V) x - (L/V) xBxint: x = xB
CMO: B = L
mostly H2O
Ex.: Open steam distillation of MeOH/H2O
VLE
y=x
1. Draw y=x line
2. Plot xD and zF on y=x
4. Draw feed line, slope = q/(q-1)
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
5. Draw top op. line, slope = L/V
• xB
•xD
•zF3. Plot xB on x-axis
•
• •
•
•
•1
6
2
5
3
4
All stages are on the column (no partial reboiler).
N = 6 NF,opt = 4
Specifications:xD = 0.9, xB = 0.07, zF = 0.5Feed is a 2-phase mixture, 50% liq.Total condenser, open steam, R = 1. Find N and NF,opt.
Column internals
• 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”
Sieve tray
Other types of trays
• 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
Valve 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
Bubble cap tray
Downcomers
Dual-pass tray
In large diameter columns, use multi-pass trays to reduce liquid loading in downcomers
Cross-flow tray (single pass)
vertical downcomer
alternates sides
• Both liquid and vapor pass through holes• Narrow operating range
Dual-flow tray (no downcomer)
Tray efficiency
effici
ency
vapor flow rate
weeping/dumping
flooding
inefficient m
ass
transfe
rexcessive
entrainment
design point
• 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: http://www.youtube.com/watch?v=QQgtcNzW9Nw&NR=1
Flooding: http://www.youtube.com/watch?v=tHOlFleAkNE
Weeping: http://www.youtube.com/watch?v=tRRxBCSuz48
Column flooding
1. jet flood (due to entrainment)
• vapor flow rate too high
2. lack of downcomer seal
• weir height below downcomer• vapor flows up downcomer
3. insufficient downcomer clearance
• weir height above downcomer• liquid backs up downcomer
• ensure bottom edge of downcomer is 1⁄2´´ below top edge of outlet weir.
Column sizing1. Calculate vapor flood velocity, uflood (ft/s)
where Csb,f is the capacity factor, from empirical correlation with flow parameter, FP
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
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:
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:
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 bottomif feed is saturated vapor, design for the top
- balance section diameters (2-enthalpy feed, intermediate condenser/reboiler)
Packed columns
• 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)
structured packing: random packing:
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