1. Continuous Column Distillation

3. Column diagram total condenser
• liquid/vapor streams inside the column flow counter-current in direct contact with each other
reflux drum
(accumulator)
enriching section
reflux
L, xR
distillate
D, xD
• all three external streams (F, D, B) can be liquids (usual case)
temperature
• for a binary mixture, the compositions xF, xD, xB all refer to the more volatile component
feed
F, xF
xR = xD
feed stage
stripping section
yB ≠ xB V L
xD ≠ K xB
• to keep the liquid flow rate constant, part of the distillate must be returned to the top of the column as reflux
boilup
V, yB
partial reboiler
• the partial reboiler is the last equilibrium stage in the system
bottoms
B, xB

4. External mass balance
TMB: F = D + B CMB: F xF = D xD + B xB for specified F, xF, xD, xB, there are only 2 unknowns (D, B)
distillate
D, xD
feed
F, xF
Define recovery and splits
B = F - D
bottoms
B, xB

5. External energy balance
assume column is well-insulated, adiabatic
EB: F hF + QC + QR
= D hD + B hB
distillate
D, xD
feed
F, xF
F, hF are known
D and B are saturated liquids
so hD, hB are also known
unknowns: QC, QR
• need another equation
bottoms
B, xB

6. Balance on condenser
1. Mass balance TMB: 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
vapor
V1, y1
reflux
L0, xR
distillate
D, xD
hD > H1
QC < 0
QR > 0

7. Splits
Sometimes 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

8. Calculating fractional recoveries
B = F – D = = 339

9. Stage-by-stage analysis Lewis-Sorel method
Consider the top of the distillation column:
vapor
V1, y1
V1, V2 are saturated vapors
L0, L1 are saturated liquids
reflux
L0, x0
distillate
D, xD
stage 1 V2 y2 L1 x1
Which streams have compositions related by VLE?
V1, L1
They are streams leaving the same equilibrium stage.
K1(T1,P) = y1/x1
Lecture 6 ends here
How are the compositions of streams V2 and L1 related?
Need to perform balances around stage 1.

10. Relationships for stage 1
vapor
V1, y1
TMB: L0 + V2 = L1 + V1 CMB: L0x0 + V2y2 = L1x1 + V1y1 EB: L0h0 + V2H2 = L1h1 + V1H1 VLE: K1(T1,P) = y1/x1
reflux
L0, x0
distillate
D, xD
stage 1 V2 y2 L1 x1
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/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.

11. 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,y2
L1,x1
stage 2
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

12. 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

13. Rectifying column 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
D, xD
replace rectifying by enriching here and below – parallel with stripping
stage j
CMB: Vj+1yj+1 = Ljxj + DxD
Vj+1,yj+1
Lj,xj
CMO: yj+1 = (L/V) xj + (D/V) xD
D = V - L
yj+1 = (L/V) xj + (1 - L/V) xD
F, xF
B, xB
Relates compositions of passing streams.

14. Lewis analysis of rectifying column
Assume CMO (Vj = Vj+1 = V; Lj = Lj-1 = L)
Need specified xD; xD = y1
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

15. 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)
y=x
VLE
•xD
= x0
(x0,y1)
op. line
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)

16. McCabe-Thiele analysis: rectifying column
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
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.

17. 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
y=x
VLE
stage 1
(x1,y1)•
•xD= x0
(x0,y1)
op. line
•(x1,y2)
(x2,y2)•
stage 2
1. Plot VLE line
2. Draw y=x line
•(x2,y3)
3. Plot xD on y=x
(x3,y3)•
stage 3
4. Plot yint = xD (1 - L/V)
•xB
Lecture 7 ends here
L/V = R/(R+1) = 2/3
yint = xD(1 - L/V)= 0.8/3 = 0.26
lowest xB possible for this op. line •
yint•
5. Draw in operating line
6. Step off stages from xD to xB
7. Count the stages
N = 3

18. Limiting cases: rectification
L/V = 0
No reflux!
Specifications:
xD = 0.8, vary R = L/D
y=x
VLE
L  0
R = L/D  0 NO REFLUX
L/V  0
stage 1
(x1,y1)•
•xD= x0
(x0,y1)
Column operates like a single equilibrium stage.
(Why bother?)
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
L/V = 1
Total reflux!
Max. distance between VLE and op. line
Max. separation on each equil. stage
Corresponds to Nmin, but no distillate!

19. Minimum reflux ratio L/V = 0 Specifications: xD = 0.8, vary R VLE •
y=x
VLE •
•xD= x0
(x0,y1)
The number of stages N required to reach the VLE-op. line intersection point is .
Increasing R = L/D
Decreasing D
Decreasing xB (for fixed N) •
This represents xB,min for a particular R.
Rmin for this xB
It also represents Rmin for this value of xB. •
xB ,min for this R
L/V = 1
0 ≤ L/V ≤ 1
0 ≤ R ≤ 

20. Ractual can be specified as a multiple of Rmin
Optimum reflux ratio
Rmin
total cost
capital cost
Ropt
operating (energy) cost
cost/lb
∞ stages
min. heat
required
external reflux ratio, R
Rule-of-thumb:
1.05 ≤ Ropt/Rmin ≤ 1.25
Ractual can be specified as a multiple of Rmin

21. Stripping column 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:
F, xF
D, xD
Vk,yk
Lk-1,
xk-1
stage k
CMB: Lk-1xk-1 = Vkyk + BxB
CMO: yk = (L/V) xk-1 - (B/V) xB
L = V + B
yk = (L/V) xk-1 + (1 - L/V) xB
B, xB

22. Graphical analysis of stripping column
equation of the operating line:
y = (L/V) x + (1 - L/V) xB
slope = L / V
always positive
y=x
VLE
op. line
(xN+1,yN+1) PR •
plotting the operating line:
y = x = (L/V) x + (1 - L/V) xB = xB
plot xB on y = x
finding the operating line slope:
•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)

23. McCabe-Thiele analysis: stripping column
Plot VLE line (yi vs. xi)
Draw the y = x line
Plot xB on y = x
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.

24. Ex.: MeOH-H2O stripping column
NEVER step over the VLE line.
(0.7,1) •
op. line
Column with partial reboiler
Specifications:
xB = 0.07,
Find N required to achieve xD = 0.55
y=x
VLE
stage 1
(xN-2,yN-2) •
stage 2
xD,max for this boilup ratio •
(xN-1,yN-1) •
•(xN-2,yN-2)
stage 3 • xD
1. Plot VLE line
(xN,yN) •
• (xN-1,yN)
2. Draw y=x line
3. Plot xB = xN+1 on y = x
4. Draw op. line • PR
• (xN,yN+1)
(xN+1,
yN+1)
5. Step off stages starting at PR •
xB= xN+1
(xN+1,yN+2)
6. Stop when you reach x = xD
7. Count the stages.

25. Limiting cases: stripping
Specifications:
xB = 0.07, vary boilup ratio
y=x
VLE
NO BOILUP
Behaves as if the column wasn’t even there.
(Why bother?)
TOTAL BOILUP
2. B  0
Operating line is y=x
TOTAL BOILUP • PR
NO BOILUP •
xB= xN+1
Max. distance between VLE and op. line
Max. separation on each equil. stage
Corresponds to Nmin. But no bottoms product!

26. Minimum boilup ratio Specifications: xB = 0.07, vary boilup ratio
yD ,max for
this boilup ratio •
y=x
VLE •
The number of stages N required to reach the VLE-op. line intersection point is .
Increasing boilup ratio
Increasing xD (for fixed N)
Decreasing B
Total boilup
This represents yD,max for a particular boilup ratio. • PR
It also represents the minimum boilup ratio for this value of yD.
No boilup •
xB= xN+1

27. McCabe-Thiele analysis of complete distillation column
Total condenser, partial reboiler
Specifications:
xD = 0.8, R = 2
xB = 0.07,
Find N required
Locate feed stage
y=x
bottom op. line
Feed enters
on stage 2
VLE
stage 1 • xB
•xD •
top op. line
stage 2 • •
1. Draw y=x line
2. Plot xD and xB on y=x
3. Draw both op. lines • PR •
4. Step off stages starting
at either end, using new op. line as you cross their intersection
NEVER step over the VLE line.
5. Stop when you reach the other endpoint
6. Count stages
7. Identify feed stage

28. Feed condition
Changing the feed temperature affects internal flow rates in the column V L
feed F
If the feed enters as a saturated liquid, the liquid flow rate below the feed stage will increase: V L
If the feed enters as a saturated vapor, the vapor flow rate above the feed stage will increase:
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:
and

29. Feed quality, q EB: rearrange: TMB: substitute: combine terms:
define:
q  mol sat’d liquid generated on feed plate, per mol feed

30. Different types of feed quality
and
saturated liquid feed q = 1
saturated vapor feed q = 0
feed flashes to form 2-phase < q < 1
mixture, q% liquid
subcooled liquid feed q > 1
- some vapor condenses on feed plate
superheated vapor q < 0
- some liquid vaporizes on feed plate

31. Equation of the feed line
rectifying section CMB:
stripping section CMB:
intersection of top and
bottom operating lines:
substitute:
Wankat gives slope as q/(q-1)
and
equation of the feed line:

32. Plotting the feed line y = x = zF
where does the feed line intersect y=x?
y=x
VLE
subcooled liq
2-phase
sat'd liq
y = x = zF
sat'd vapor
•zF
Wankat gives slope as q/(q-1)
superheated vap
feed type q slope, m
sat'd liquid q=1 m = 
sat'd vapor q=0 m = 0
2-phase liq/vap 0<q<1 m < 0
subcooled liq q>1 m > 1
superheated vap q<0 0<m<1

33. 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.
y=x 1 6 2 5 3 4 • xB
•xD
•zF
N = 6 + PR • • • • •
1. Draw y=x line
Operating lines intersect on stage 4. This is NF,opt.
2. Plot xD, xB and zF on y=x •
3. Draw feed line, slope = -0.5
4. Draw top op. line, slope = L/V = 0.5
Optimum feed stage has intersection point underneath the ‘step’
We can independently specify only 2 of the following 3 variables: R, q, V/B (usually: R, q).
5. Draw bottom op. line (no calc. required) PR •
6. Step off stages starting at either end, using new op. line as you cross the feed line.
Using a non-optimal feed location reduces separation.

34. Feed lines in rectifying/stripping columns
rectifying column
stripping column • zF •
•xD •
• yD • • •
top operating
line • • • • • •
•zF • PR •
bottom operating line
May be slightly incompatible with previous description of these column endpoints - check
•xB • xB
total condenser, no reboiler
sat’d vapor feed, liquid bottoms
F and B are passing streams
partial reboiler, no condenser
sat’d liquid feed, vapor distillate
F and V are passing streams

35. Design freedom Fixed q. Vary R: Fixed R. Vary q: heat the feed • xB
•xD
•zF
Rmin • xB
•xD
•zF
pinch point
decrease R
qmin
pinch point
choice of R dictates required boilup ratio.
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.

36. Another type of pinch point
Ethanol-water
xD = 0.82, xB = 0.07
zF = 0.5, q = 0.5
Find Rmin
y=x • xB
•xD
•zF
pinch point
1. Draw y=x line
VLE
2. Plot xD, xB and zF on y=x
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.

37. Additional column inputs/outputs
Column with two feeds:
bottoms
B, xB
distillate
D, xD
Column with three products:
distillate
D, xD
bottoms
B, xB
feed
F, z V L V L
feed 2
F2, z2, q2
side-stream
S, xS or yS
z2 > z1
and/or
q2 > q1 V´ L´ V´ L´
side-streams must be saturated liquid or vapor
feed 1
F1, z1, q1 V L V L
Each intermediate input/output stream changes the mass balance, requiring a new operating line.

38. 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
y=x
VLE 1 PR 2 5 3 4 • xB
•xD
•z2
•z1 • • • •
1. Draw y=x line
2. Plot xD, z1, z2 and xB on y=x •
3. Draw both feed lines
•z1 = z2
4. Draw top op. line, slope = L/V
5. Calculate slope of middle operating line, L´/V´, and draw middle operating line •
Optimum location for feed 1 is stage 5.
Optimum location for feed 2 is stage 3.
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

39. Slope of middle operating line
2-feed mass balances:
TMB: F2 + V´ = L´ + D
CMB: F2z2 + V´yj+1 = L´xj + DxD
feed 2
F2, z2, q2
D, xD
middle operating line equation:
y = (L´/V´)x + (DxD - F2z2)/V´
obtain slope from:
L´ = F2q2 + L = F2q2 + (R)(D)
V´ = L´ + D – F2 V´ L´
stage j
side-stream  feed-stream with –ve flow rate sat’d liq y = x = xS
sat’d vapor y = x = yS
side-stream
S, xS or yS
D, xD
Not finished
side-stream mass balances:
TMB: V´- L´= D + S
CMB: V´yj+1 - L´xj = DxD + SxS V´ L´
stage j
middle operating line equation:
y = (L´/V´)x + (DxD + SxS)/V´

40. McCabe-Thiele analysis of side-streams
Saturated liquid side-stream, xs = 0.64
Saturated vapor side-stream, ys = 0.73
VLE
y=x •
y=x • xB
•xD
•xS •z • xB
•xD
•yS •z • • • •
VLE
Side-stream must correspond exactly to stage position.

41. Partial condensers y=x •
A partial condenser can be used when a vapor distillate is desired: PC 1 2
•yD • •
D, yD V L
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

42. Total reboilers
A total reboiler is simpler (less expensive) than a partial reboiler and is used when the bottoms stream is readily vaporized:
y=x TR N
N-1 •
B, xB V L
stage N
V, yB •
•xB,yB
A total reboiler is not an equilibrium stage.

43. 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:
Can also define Murphree liquid efficiency:
where yn* is the equilibrium vapor composition (not actually achieved) on stage n:
yn* = Kn xn
xj* = yj / Kj

44. 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.
y=x • • xB
•xD •z 2 8 3 6 4 5 1 7 •   •  •  •  • 
1. Draw y=x line • 
2. Plot xD, z, and xB on y=x •
3. Draw feed line 
4. Draw top op. line, slope = L/V
5. Draw bottom operating line (no calc. required)
N = 8 + PR PR •
NF,opt = 6
6. Find partial reboiler
7. Step off stages, using EMV to adjust vertical step size.
8. Label real stages.
To use ELV, adjust horizontal step size instead.

45. Intermediate condensers and reboilers
Intermediate condensers/reboilers can improve the energy efficiency of column distillation:
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
by decreasing the cooling that must be supplied at the top of the column, providing part of the cooling using an intermediate condenser instead
- use a smaller (cheaper) cooling element at the top of the column, and/or a higher temperature coolant for the intermediate condenser
distillate
D, xD
bottoms
B, xB
feed
F, z V L V´ L´
S, xS
intermediate reboiler
yS = xS
V´´
L´´ V L
Each column section has its own operating line.

46. 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. V2 L1
L0, x0
V1, y1
D, xD
stage 1 c
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 L1 = L0 + c
CMO is valid below stage 1. Find L/V = L1/V2?
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
q0  quality of reflux
Subcooled reflux causes L/V to increase.
Superheated boilup causes L/V to increase.
where L0/V1 = (L0/D)/(1 + L0/D) = R/(R + 1)

47. 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
Vj+1,
yj+1
Lj, xj
TMB: V + B = L + S
usually 0
CMB: V yj+1 + B xB = L xj + S yS
mostly H2O
bottoms
B, xB
CMO: B = L
S, yS
B, xB
Operating line equation:
y = (L/V) x - (L/V) xB
xint: x = xB

48. 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.
y=x
VLE 1 6 2 5 3 4
•xD
•zF • • •
1. Draw y=x line •
2. Plot xD and zF on y=x •
3. Plot xB on x-axis
4. Draw feed line, slope = q/(q-1)
5. Draw top op. line, slope = L/V
All stages are on the column (no partial reboiler). •
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
N = 6 NF,opt = 4 • xB

49. 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”

50. Other types of trays Valve tray Bubble cap 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
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

51. Downcomers Dual-flow tray (no downcomer)
• Both liquid and vapor pass through holes
• Narrow operating range
In large diameter columns, use multi-pass trays to reduce liquid loading in downcomers
Cross-flow tray (single pass)
vertical downcomer
alternates sides
Dual-pass tray
Add pipe downcomer
four-pass tray

52. Tray efficiency
design point 
excessive entrainment
efficiency
inefficient mass transfer
flooding
weeping/dumping
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

53. Column distillation videos
Normal column operation: Flooding: Weeping:

54. Column flooding 1. jet flood (due to entrainment)
3. insufficient downcomer clearance
• weir height above downcomer
• liquid backs up downcomer
2. lack of downcomer seal
• weir height below downcomer
• vapor flows up downcomer
• vapor flow rate too high
• ensure bottom edge of downcomer is 1⁄2´´ below top edge of outlet weir.

55. Column sizing 1. 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

56. Tray spacing

57. 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 bottom
if feed is saturated vapor, design for the top - balance section diameters (2-enthalpy feed, intermediate condenser/reboiler)

58. 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)