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: