CHAPTER 12. PACKED COLUMNS
Learning Objectives and Expectations
To be able to calculate pressure drops of single phase flows through packed
columns (Carmen and Ergun)
To be able to determine the gas phase pressure drop in two phase (gas +
liquid) countercurrent flows through packed columns.
To be familiar with the main features of packed columns such as liquid
distributors, gas distributors, support and holddown plates, mist eliminators.
To be familiar with different packing types rings, saddles, etc.
To be familiar with important properties of packings such as size, S , F etc.
B P
To be able to use a desired pressure drop to determine absorber column
diameter.
To be able to determine the height of the packed column.
To be able to design packed columns safely and cost effectively.
To be able to produce packed column specification sheets and equipment
sketches.
To be familiar with the rules of thumb to select packed or plate columns
Chapter 12 Packed Columns 1
CHAPTER 12. PACKED COLUMNS
1. FLOW IN PACKED COLUMNS
Single Phase Flow - liquid or gas, but not both.
Two Phase Flow - both liquid and gas, counter current or co-current
(will discuss counter-current flow only here)
2. PACKED COLUMN – SINGLE PHASE FLOW
Widely used in:
- Catalytic reactors – solid-gas reaction
Adsorption column
- Ion exchange units – solid-liquid interaction
3. GENERAL OPERATIONS – SINGLE PHASE FLOW
- Gas or liquid is passed through packing
- Fluid is the continuous phase
- Interaction between fluid and solid is required generally
- Operation is flow rate limited by rate of reaction and ∆P
- ∆P through packing limited by packing size, voidage and mean
channel diameter
Chapter 12 Packed Columns 2
4. PACKED COLUMN – TWO PHASE FLOW
Widely use in
- Distillation column
- Absorber
- Humidifier (TEG), etc
5. GENERAL OPERATIONS – TWO PHASE FLOW
- Provides good contact between liquid and vapour phases
- Co-current flow, less commonly used
- Counter-current flow, widely used
- Liquid is passed through the packing via a distributor, usually from
the top
- Gas is passed through the packing, usually from the bottom
- Distributed liquid trickles down the packed bed
- Liquid on the particles in bed is exposed to the gas
- To facilitate absorption, stripping, and separation
Video
- Particles in bed usually have a large specific surface area
- Efficient mass transfer occurs at the gas/liquid interface
Chapter 12 Packed Columns 3
Typical absorption process
Cold
Hot
Animation at http://www.co2crc.com.au/misc/Schematic_1_animation/absorption_n_desorption_animation.html
6. TYPICAL FEATURES
- Shell (pressure vessel or non-
pressure vessel)
- Bottom support for packing and gas
distributor
- Packings – random and/or structured
- Liquid distributor, gas distributor
- Inlet and outlet nozzles
- See Fig. 6.27, Treybal
Chapter 12 Packed Columns 5
Column internals
Packing material, plus
Liquid inlet systems
Liquid & vapour distributors
Liquid collecting devices Video
Packing supports
Good info at manufacturer
www.sulzechemtech.com
GREEN, D. W. & PERRY, R. H. (eds.) (2008).
6
Perry's chemical engineers' handbook, New
York: McGraw-Hill.
Packed columns – random
packings
Raschig rings
see more images at
www.tower-packing.com
Metal pall rings
VSP Inner arc ring
7
Structured packings
www.sulzerchemtech.com
MellapakTM
www.sulzerchemtech.com
Grids
7. PARAMETERS LIMITING THE OPERATING CONDITIONS OF A
PACKED COLUMN
a. Packing Size and Voidage
- This governs the mean channel diameter available for fluid flow
- This in turn affects the flow velocity and pressure drop across the
packing
b. Column Diameter
- Overall volumetric flow rates are determined by cross sectional area
- Must allow sufficient vapour and liquid flow rates without having
excessive pressure drop or high velocities
c. Pressure Drop
- Determined by voidage and vapour and liquid flow rates
- An optimum ∆P allows the tower to operate economically
- An optimum ∆P also allows sufficient liquid hold up and efficient
mass transfer
- Excessive ∆P can lead to flooding of tower
d. Packing Surface Area
- This governs the rate of reaction and rate of mass transfer
Chapter 12 Packed Columns 9
8. PRESSURE DROP IN A COLUMN
8.1 Single Phase Flow
1. Carmen’s Equation
- For both laminar and turbulent flow.
R ∆P 1 ε3
1 = − = 5Re−1+ 0.4Re−0.1
ρu2 Z ρu2 S(1−ε) 1 1
1 c
u ρ
Where Re = c Curve A
1 S(1−ε)µ
u
u = c
1 ε
S = specific surface area of the particles
S = surface area per unit volume of packed bed
B
u = mean velocity through the column, (superficial velocity)
c
difference
u = mean velocity through the pores
1
Chapter 12 Packed Columns 10
S = specific surface area of the particles
S = surface area per unit volume of packed bed
B
u = mean velocity through the column, (superficial velocity)
c
u = mean velocity through the pores
1
ε = e = voidage
R
1 =
Friction Factor
ρu2
1 What is its unit? pressure
drag force
R is the component of in the direction of motion
1 unit area of particle surface
R
1 =
is related to Re by Carmen as a continuous curve (Curve A)
ρu2 1
1
See Fig. 4.1, C&R, Vol. 2, next page
Chapter 12 Packed Columns 11
Carman
Curve A
Ergun
Curve C
Note: e = ε = voidage
Chapter 12 Packed Columns 12
2. Ergun’s Correlation
- Ring Packing
- For streamline as well as turbulent flow
−∆P 150(1−ε)2µu ρu2 ( 1−ε)
= c +1.75 c
Z
ε3 d2 ε3
d
p p
( ) ( )
6 1−ε 6 1−ε
d = =
p
a S
p B
Treybal C&R
S = a = specific surface area per unit volume of packed bed m2/m3
B p
d = diameter of the particles (packings)
p
u = fluid velocity, based on entire tower diameter
c
∆P = pressure drop
Z = packing height
Chapter 12 Packed Columns 13
6 S
For a sphere S = = B
d 1−ε
p
R − ∆P ε3
1 = = 4.17 Re−1+ 0.29
ρu2 ρu2
SZ
( 1−ε) 1
1 c
Curve C
u ρ
Re = c
Where 1 ( )
S 1−εµ
Chapter 12 Packed Columns 14
8.2 Two phase flow – gas and liquid
- Consider counter current flow only
- Predictions are not very accurate
- Typical pressure drop in random packing
- Fig. 6.33, Treybal, pg. 12 and Fig. 11.44, C&R, Vol. 6, pg 15.
Chapter 12 Packed Columns 15
For packed column
Design considerations: Pressure drop and flooding
Gas outlet
Liquid inlet Some flooding description
•A visual build-up of liquid on the upper
surface of the packed bed
• A rapid increase in liquid hold-up with
increasing gas rate
• Formation of a continuous liquid phase above
the packing support plate
• A considerable entrainment of liquid in
the outlet vapour
Liquid outlet Gas inlet
• Filling of the voids in the packed bed with liquid
www.see.ed.ac.uk
∆P increase with liquid load
Chapter 12 Packed Columns 17
- Slope of line for dry packing 1.8 to 2.0
- Turbulent flow for most practical gas velocities
- ∆P increase with increasing liquid rate because void area for
flow of gas is reduced, a thicker liquid layer is found on packing
surface.
- Below A, loading point, the liquid loading is reasonably constant
with increasing gas velocity, although it increases with liquid
rate. Gas-continuous phase, liquid-dispersed phase.
- Between A and B, liquid holdup increases rapidly with gas rate,
the free area for gas flow decreases. The pressure drop rises
more rapidly with gas velocity. Slope increase is apparent, this
is known as loading. Liquid is loaded on particle surface.
- Above B – liquid may fill the tower starting from the bottom. ∆P
increases dramatically gas-dispersed phase, liquid –
continuous. The tower is flooded and impractical to operate.
- Fig. 11.44, C&R, Vol. 6, gives the condition when flooding may
be expected, and the ∆P in the column before the flooding
point.
Chapter 12 Packed Columns 18
Figure 11.44 correlates the liquid and vapour flow rates, system
physical properties and packing characteristics, with the gas mass
flow-rate per unit cross-sectional area; with lines of constant
pressure drop as a parameter.
The term K on Figure 11.44 is the function:
4
0.1
µ
( )
2
13.1V * F L
w p ρ
(11.118, C&R, Vol. 6)
K = L
4 ( )
ρ ρ −ρ
V L V
*
V = G* = gas mass flow-rate per unit column cross-sectional area, kg/m2s
w
F = packing factor, characteristic of the size and type of packing, m-1
p
See table 11.3, C&R Vol 6 or Table 4.3 C&R vol 2
µ
= liquid viscosity, Ns/m2
L
ρ ,ρ
= liquid and vapour densities, kg/m3
L V
Chapter 12 Packed Columns 19
Chapter 12 Packed Columns 20
9. DATA REQUIRED FOR DESIGN
1. Feed flow rate and composition
2. Extent of separation required
3. Equilibrium and enthalpy data
4. Operating conditions – T&P
5. Physical property data of fluids
6. Physical properties of packing
10. DESIGN PROCEDURE (ABSORBER)
1. Select the type and size of packing
2. Determine column height (process design)
3. Determine column diameter (hydraulic considerations)
- Liquid and vapour flow rates
4. Select column internal features (mechanical design)
- Packing support
- Liquid distributor
- Inlet and outlet ports etc.
Chapter 12 Packed Columns 21
11. PACKINGS
- Packing characteristics preferred:
A large interfacial surface between liquid and gas,
∴ S is large
B
S = specific surface area
B
( )
2
exposed surface area m
S = ( )
B 3
volume of packing m
- Have an open structure to allow flow of liquid and vapour.
- Be chemically inert
- Promote liquid distribution and wetting.
- Have structural strength for installation
- Low cost
Chapter 12 Packed Columns 22
12. TYPES OF PACKINGS
Random packings:
Raschig rings
Pall rings
Berl saddles
Intalox saddles
Tellerettes
Tripaks
Structured packings: Lots of
newly developed products
Walas
Chapter 12 Packed Columns 23
13. DATA FOR PACKING
- Nominal size
- Bulk density (kg/m3)
- Specific surface area, S , m2/m3
B
- Packing factor, F S /ε3 but more accurate because it is based
P ≈ B
on performance rather than calculation.
- Voidage
- Table 4.3, C&R, Vol. 2, next two pages
Chapter 12 Packed Columns 24
Chapter 12 Packed Columns 25
Chapter 12 Packed Columns 26
14. PACKING SIZE AND TYPE
- Consult the supplier
- Traditionally made of ceramics and metals
- Plastic ones are now readily available with good structural strength
Recommended sizes (C&R, Vol 6)
Column Diameter Packing Size
<0.3m <25mm
0.3-0.9m 25-38mm
>0.9m 50-75mm
- A compromise between surface area for mass transfer and voidage
for fluid flow.
- Due to deformability, maximum packing depth unsupported
Plastic packing: 4-5 m max.
Metal: ∼ 7m max. (Walas)
- Key feature is S
B
Chapter 12 Packed Columns 27
15. COLUMN INTERNALS
Structural Support
- Cross beams to support weight of packing and liquid
- Gas injection base and support plate
- See Fig. 6.30, Treybal, next page
Liquid Distributors
- Several types available
- See Fig. 4.12, C&R, Vol 2 and Fig. 6.31 & 6.32
Treybal, pg. next three pages
Chapter 12 Packed Columns 28
Support plate
Chapter 12 Packed Columns 29
Chapter 12 Packed Columns 30
Chapter 12 Packed Columns 31
Hold Down Plate
- To prevent fluidization of packing particles, packing restrainer to
hold the packing materials
Entrainment Eliminator
- Mist eliminator
- Knitted mesh, 100 mm thick, wire or polyethylene
Liquid Redistributor
- The packing density (number of particles per unit volume) is less
near the wall.
- This leads to tendency of the liquid to segregate towards the wall
- Gas channels in the centre of tower
D
- Smaller packing tends to reduce such tendency ≥ 1 5 is
d
recommended. p
- Redistribution every 5 to 10 tower diameter.
Chapter 12 Packed Columns 32
16. PROCESS DESIGN FOR A PACKED ABSORBER
1. Collect equilibrium data – Prepare X-Y diagram
X = mole ratio of component in the liquid
Y = mole ratio of component in the gas
x = mole fraction of component in the liquid
y = mole fraction of component in the gas
2. Specify extent of separation
∴ Specify Y and Y
1 2
Specify purity of absorbing liquid
∴ Specify X
2
3. Calculate minimum slope
L
s min
Slope =
min
G
s
Chapter 12 Packed Columns 33
Review slides
Liquid and Gas molar L G
s s
Flow rates L G
2 2
x y mole fractions
2 2
Overall mass balance
2
V y + L x = V y + L x
1 1 2 2 2 2 1 1
Note V = G meaning vapour or gas
1
(water) L G (air)
s s
(water +SO ) L G (air + SO )
2 1 1 2
x y
1 1
mole fractions
Chapter 12 Packed Columns 34
Review slides
Often convenient to write in terms of mole
ratio (instead of mole fraction)
V : moles of phase V on solute free basis/ time unit
s
L : moles of phase L on solute free basis/ time unit
s
y : mole fraction component A in V
1 1
x : mole fraction component A in L
1 1
+ = +
V Y L X V Y L X
s 1 s 2 s 2 s 1
y x L Y − Y
where Y = , X =
s = 1 2
1 − y 1 − x
−
V X X
s 1 2
Y X
and y = , x =
1 + Y 1 + X
35
Review slides 100
YA vs yA
yA vs yA
10
n
1
Y = A
A
n
B
0.1
0.01
0.01 0.1 1
n
y = A
A
n + n
A B 36
Minimum operating line
Review slides
Real operation line
(X , Y )
1,real 1
Slope =
Ls(min)/Vs
(X *, Y )
1 1
Equilibrium
curve Y*=f(X*)
L −
Y Y
(X2, Y2)
s = 1
−
V X X
s 1
37
In class quiz
Which way will component A transfer?
1. Vapour to liquid
2. Liquid to Vapour
3. There is no mass transfer
38
4. Calculate operating slope
Slope = 1.5 slope
op min
L = 1.5 L
s op s min
5. Calculate X This is actually the X
1 1,real
G (Y – Y ) = L (X – X )
s 1 2 s 1 2
6. Draw operating line on X-Y diagram
7. Determine the number of theoretical trays
Chapter 12 Packed Columns 39
17. HEIGHT OF THE PACKED BED
It is more convenient to use mole fraction (x and y) for this part of the
calculations but if you have the Y- X equilibrium equation it is totally
fine to use (X, and Y).
The equilibrium and operating curves on the X-Y (mole ratio) diagram
can be converted to the corresponding curves on the x-y (mole
fraction) diagram.
The following equations will be useful for the conversion:
x
X =
1− x
y
Y =
1− y
Chapter 12 Packed Columns 40
Z = H N
toG toG
G dy
y
Z = ∫ 1
use this if K C > K P where K and K are
K aP y y − y L T G T L G
2 the mass transfer
G T e
coefficients,
Mass transfer resistance lies in the vapour phase. C and P are the
T T
concentration and
Z = N H partial pressure of
toL toL
the component and
L x dx a = the effective
Z = ∫ 1
use this if K C < K P wetted area.
K aC x x − x L T G T
2
L T e
Mass transfer resistance lies in the liquid phase
See H & M 1 notes or Pong notes for more details
Chapter 12 Packed Columns 41
18 DIAMETER OF PACKED COLUMN
Use Fig. 11.44, C&R, Vol 6, pg. 15
L* ρ
w V ≡ x − axis
1. Calculate
G* ρ
w L
2. Read the K value on the y axis using the flooding line
4
3. Pick a value for ∆P below the flooding line
TYPICAL ∆P
mm water
For absorber and stripper 15-50
m packing
mm water
Distillation 40-80
m packing
Chapter 12 Packed Columns 42
K
4 flood
K
4 oprt
Chapter 12 Packed Columns 43
4. Calculate V * using Eq. 11.118 (see earlier pg these notes, C&R Vol 6)
w
0.1
µ
( )
2
13.1V * F L F , packing factor
p
w p ρ
K = L
4 ( )
ρ ρ −ρ
V L V
*
V
5. Calculate percentage of Flooding for
w
K α V *2
4 w
(Eq. 11.118, C&R, Vol. 6)
K = CV *2
4 w
*
K V
4operating
=
Woperaing
≅ 60 − 80% flooding
*
K V
4 flooding W flooding
Chapter 12 Packed Columns 44
6. Calculate column cross sectional area
Mass flow rate kg / s
= = m2
2
Mass flux kg / m s
7. Calculate column diameter
4
Diameter = Area
π
May need rounding off
8. Check packing size to diameter ratio
Chapter 12 Packed Columns 45
19. WETTING RATE OF PACKING SURFACE
Volumetric liquid rate per cross-sectional area
Wetting Rate =
Packing surface area per unit volume
m3/m2s m2
Liquid flux
W.R. = = =
m2/m3
s packing surface area
1. Morris and Jackson (1953)
For rings between 25-75 mm, W.R. = 2.0 x 10-5 m3/ms
For rings larger packings W.R = 3.3 x 10-5 m3/ms
2. Norman (1961) determined on a mass basis
kg
Recommended liquid rate = > 2.7
m2s
3. A nomograph is presented
See Fig. 4.23, C&R, Vol. 2, next page
Chapter 12 Packed Columns 46
>2.0 x 10-5 m3/ms
Chapter 12 Packed Columns 47
20. DESIGN PROCEDURE (PACKED DISTILLATION COLUMN)
1. Prepare equilibrium curve
2. Perform material balance to get liquid and gas flow rate, L, G, L, G
3. Determine stream compositions, X , X , X
D W F
4. Determine location of maximum vapour and liquid rates
5. Select a packing material.
6. Determine height of packed column
7. Calculate column diameter using area of maximum vapour and
liquid flow
8. Design column internal and support
Chapter 12 Packed Columns 48
21. PLATE COLUMN VS PACKED COLUMN
Plate Column Packed Column
Can handle a wide range of Can handle a smaller range, not
liquid and vapour flow rates suitable for low liquid flow rate
Generally good liquid distribution Poor liquid distribution and poor
wetting can be a problem
Easier to install ports for side Multiple trays of packing may be
streams needed when side streams are
withdrawn or input.
Easier to make provision for Installing a cooling coil is more
cooling difficult
Higher liquid hold up Much lower liquid hold up
Equipment cost higher for Cheaper for handling corrosive
corrosive liquids liquids
Not suitable for foaming liquid – Handles foaming liquid better
reduces disengagement space
Higher pressure drop Pressure drop generally lower
Chapter 12 Packed Columns 49
PACKED ABSORBTION COLUMN WORKED EXAMPLE
Given: SO in air is to be absorbed in water
2
Feed = 3800 kg/hr (air + SO )
2
8 mole% SO
2
Specification = 95% SO removal
2
Temp = 20oC Pressure = 1 atm
Air ρ = 1.177 kg/m3
Water ρ = 1000 kg/m3
1. Provide specifications for the packed absorber
2. If 1m of packing is used above the liquid inlet as an entrainment
separator, determine overall ∆P across the entire unit
Chapter 12 Packed Columns 50
Solubility data
Obtain equilibrium data from
literature (shown on the right) and
examine the data.
wt% SO partial pressure
2
in solution mm Hg
If the solubility of SO in water is
2
0.05 1.2
too low, then only limited
0.1 3.2
separation is possible.
0.15 5.8
Try other methods of separation if 0.2 8.5
necessary. 0.3 14.1
0.5 26
0.7 39
1 59
1.5 92
Also need gas side properties
Chapter 12 Packed Columns 51
Liquid and Gas molar L G
s s
Flow rates L G
2 2
x y mole fractions
2 2
2
1
(water) L G (air)
s s
(water +SO ) L G (air + SO )
2 1 1 2
x y mole fractions
1 1
Chapter 12 Packed Columns 52
(a) GAS SIDE PROPERTIES
M.W. (air) = 29 kg / kmole
M.W. (SO ) = 64 kg / kmole
2
Avg. gas M.W. = (0.08) (64) + (0.92) (29)
= 31.8 kg / kmole
Gas flow rate, G , (air + SO )
1 2
(3800 kg/hr)
G =
1 (31.8 kg/kmole) (3600 s/hr)
= 0.0332 kmole / s
Air flow rate, G , (air only)
s
kmole
G = (0.92) (0.0332 kmole / s ) = 0.0305
s s
Chapter 12 Packed Columns 53
Mass gas flow rate, G’
1
(3800 kg/hr)
G’ = = 1.056 kg/s, air + SO
1 (3600 s/hr) 2
Mass air flow rate, G’
s
kmole kg
G’ = (0.0305 ) (29 )
s s kmole
= 0.885 kg/s
Mass SO input flow rate, G’ ,
2 1 SO2
G’ = 1.056 – 0.885 kg/s
1,SO2
= 0.171 kg/s
Chapter 12 Packed Columns 54
Mole fraction of SO in feed = y = 0.08
2 1
Partial pressure of SO in feed, P
2 1
P = (0.08) (760 mm Hg) = 60.8 mm Hg
1
Partial pressure of SO in exit gas after 95% recovery, P
2 2
P = (0.05) (60.8 mm Hg) = 3.04 mm Hg
2
3.04 mm Hg
y = = 0.0040
2
760 mm Hg
or Y = (1- 0.95) Y = y / (1 – y )
2 1 2 2
Chapter 12 Packed Columns 55
x = 0 y = 0.0040
2 2
Water Air + trace SO
2
Y = 0.00402
2
2
95%
removal
1
Air + 8% SO
2
F = 3800 kg/hr
y = 0.08
1
Y1 = 0.087
Chapter 12 Packed Columns 56
Note that:
x, y ≡ mole fraction
X, Y ≡ mole ratio
and
x
X =
1− x
y
Y =
1− y
y = 0.08 → Y = 0.087
1 1
y = 0.0040 → Y = 0.00402
2 2
Chapter 12 Packed Columns 57
Equilibrium data
Mole Mole Mole Mole
Solubility Data fraction fraction ratio ratio
SO % w/w Partial Pressure in liquid in gas in liquid in gas
2
Solution gas, mm Hg x y* X Y*
0.05 1.2 1.407 x 10-4 1.579 x 10-3 1.407 x 10-4 1.581 x 10-3
0.10 3.2 2.815 x 10-4 4.211 x 10-3 2.815 x 10-4 4.228 x 10-3
0.15 5.8 4.223 x 10-4 7.631 x 10-3 4.225 x 10-4 7.690 x 10-3
0.20 8.5 5.633 x 10-4 1.118 x 10-2 5.636 x 10-4 1.131 x 10-2
0.30 14.1 8.454 x 10-4 1.855 x 10-2 8.463 x 10-4 1.890 x 10-2
0.50 26.0 1.411 x 10-3 3.421 x 10-2 1.413 x 10-3 3.542 x 10-2
0.70 39.0 1.979 x 10-3 5.132 x 10-2 1.983 x 10-3 5.409 x 10-2
1.00 59.0 2.833 x 10-3 7.763 x 10-2 2.841 x 10-3 8.417 x 10-2
1.50 92.0 4.265 x 10-3 1.211 x 10-1 4.283 x 10-3 1.377 x 10-1
0.05gmSO 0.05
2
MW 64
x = = y = 0.08 → Y = 0.087
0.05gmSO 99.95gmH O 0.05 99.95 1 1
2 + 2 +
MW MW 64 18
x =1.407 × 10−4 y = 0.0040 → Y = 0.00402
2 2
1.2mmHg of SO
y = 2 =1.579 × 10−3
760mmHg total
Chapter 12 Packed Columns 58
xy
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
x
Chapter 12 Packed Columns 59
y
XY
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
X
Chapter 12 Packed Columns 60
Y
(b) Determine No. of stages
Draw equilibrium curve
L
s
Draw operating line
G
s min
Bottom of tower, Y = 0.087
1
Top of tower, Y = 0.00402
2
X = 0 (water)
2
Chapter 12 Packed Columns 61
XY
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
X
Chapter 12 Packed Columns 62
Y
Y =
1
0.087
Y =
2
0.00402
X = 0.0
2
XY
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
X
Chapter 12 Packed Columns 63
Y
Y =
1
0.087
slope =
(L/G)
min
Y =
2
0.00402
X = 0.0 Then X = 0.00292
2 1
If you use (L/G) then column will be infinitely tall
min
L −
Y Y
0.087 - 0.004
s = 1
(slope) = = 28.4
min 0.00292 - 0
−
V X X
G
Use 1.5 of (slope) s 1
s
min
Operating slope = (1.5) (28.4) = 42.61
0.087 – 0.004
And 42.61 =
X - 0
1
⇒ X = 0.00195
1
Operating
x = 0.00194
1
Draw actual operating line,
X = 0.00195, Y = 0.087
1 1
Moving point 1 away from the
equilibrium line
Chapter 12 Packed Columns 64
XY
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
X
Chapter 12 Packed Columns 65
Y
Y =
1
0.087
Operating
Line
Y =
2
0.00402
X = 0.0
Operating X = 0.00195
2
1
Get N by graphical integration
toG
dy 1 1− y
y
N = ∫ 1 + ln 2
toG
y y − y * 2 1− y
2
1
Chapter 12 Packed Columns 66
Operation line and equilibrium line Graphic integration: 1/(Y – Y*) as a
function of Y
xy
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0.0045
x
Chapter 12 Packed Columns 68
y
Draw Operating line on xy chart
y =
1
0.080
Operating
Line
y =
2
0.0040
x = 0.0 Operating x = 0.00194
2 1
Graphical Method
0.14
0.12
0.1
0.08
0.06
0.04
0.02
0
0 0.0005 0.001 0.0015 0.002 0.0025 0.003 0.0035 0.004 0
Chapter 12 Packed Columns 69
x
y
Slope xy = (0.08-0.0040)/(0.00194-0) = 39.175
y = 39.175 x + 0.0040
y
y*
The slide maker switched it to x-y system.
Actually X-Y is much more convenient in graphical method
y obtained from linear equation (operation line)
y* obtained graphically (equilibrium curve)
x y y* 1/(y-y*)
0.0000 0.0040 0.0000 250.00
0.0001 0.0079 0.0010 144.56
0.0002 0.0118 0.0023 104.88
0.0004 0.0197 0.0070 78.93
0.0006 0.0275 0.0118 63.67
0.0008 0.0353 0.0165 53.08
0.0010 0.0432 0.0220 47.23
0.0012 0.0510 0.0278 43.08
0.0014 0.0588 0.0330 38.69
0.0016 0.0667 0.0390 36.13
0.0018 0.0745 0.0450 33.88
0.0019 0.0800 0.0490 32.26
Chapter 12 Packed Columns 70
Determine the value of the integral graphically plot
1/(y-y*) vs y
graphical integration
250
200
150
100
50
0
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
y
Chapter 12 Packed Columns 71
)*y-y(/1
Determine the value of the integral graphically plot
1/(y-y*) vs y
graphical integration
250
200
150
100
50
0
0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
y
Chapter 12 Packed Columns 72
)*y-y(/1
y range dy 1/(y-y*) mean dy/(y-y*)
0.07 - 0.08 0.01 34 0.34
0.06 - 0.07 0.01 37 0.37
0.05 -0.06 0.01 41 0.41
0.04 -0.05 0.01 45 0.45
0.03 -0.04 0.01 54 0.54
0.02 -0.03 0.01 68 0.68
0.01 -0.02 0.01 92 0.92
0.004 -0.01 0.006 165 0.99
Σ dy/(y-y*) 4.70
1 1− y Integrate and get N
toG
ln 2 = 0.039
2 1− y dy 1 1− y
y
1 N = ∫ 1 + ln 2
toG
y y − y * 2 1− y
1
for y =0.004, 1
2
and y = 0.08
1 N = 4.74 stages
toG
Chapter 12 Packed Columns 73
(c) Liquid Side Properties
(d) Select Packing Type
L Y −Y
s = 1 2 = operating slope
G X − X
s 1 2 TRY Intalox saddles, ceramic
L = (42.61) (0.0305 kmole/s)
Size 1 ½“ (38 mm)
s
L = 1.30 kmole/s
F = 170 m-1 packing factor
s
p
L’ = (1.30 kmole/s)(18 kg/kmole)
S = 194 m2/m3 surface area
s
B
L’ = 23.39 kg/s
Table 4.3
s
Mass of liquid leaving column, L’
1
L’ = water + SO removed
1 2
L’ = 23.39 + (0.95) (0.171) = 23.55 kg/s
1
Chapter 12 Packed Columns 74
Chapter 12 Packed Columns 75
Chapter 12 Packed Columns 76
(e) Determine % flooding
Use Fig. 11.44, need to get F
LV
L* ρ
F = V
LV
G * ρ
L
Note:
'
L
kg kg
1
L* 2 L'
m s A s
= = 1
Where area is not yet known
G* kg G' G' kg
1 1
m2 s s
A
23.55 1.177
F = = 0.765
LV
1.056 1000
Design for a ∆P of 21 mm water/m packing, using Fig. 11.44 on next page
K (flooding) = 0.75
4
K (operating) = 0.32
4
Chapter 12 Packed Columns 77
K =
4
0.75
K =
4
0.32
F = 0.765
LV
Chapter 12 Packed Columns 78
13.1G*2F µ 0.1
K = p L F in m−1 Note G* = V w*
4 ρ ( ρ −ρ ) ρ p
V L µ L
note K αG*2
4
K ρ (ρ −ρ ) ρ 0.1
G* = 4 V L V
L
13.1 F µ
p L
1/2
( 0.32)( 1.177)( 1000−1.177)
1000
0.1
G* =
( operating)
( 13.1)( 170) 1 × 10−3
G* = 0.819 kg / m2s ( operating)
1/2
( )( )( ) 0.1
0.75 1.177 1000−1.177 1000
G* = flooding
( 13.1)( 170) 1 × 10−3
G* =1.254 kg / m2s( flooding)
Chapter 12 Packed Columns 79
Review slide
4. Calculate V * using Eq. 11.118 (see earlier pg these notes, C&R Vol 6)
w
0.1
µ
( )
2
13.1V * F L
w p ρ
K = L
4 ( )
ρ ρ −ρ
V L V
*
V
5. Calculate percentage of Flooding for
w
K α V *2
4 w
(Eq. 11.118, C&R, Vol. 6)
K = CV *2
4 w
*
K V
4operating
=
Woperaing
≅ 60 − 80% flooding
*
K V
4 flooding W flooding
Chapter 12 Packed Columns 80
( )
*
G operating
% flooding =
( )
*
G flooding
2
0.819 kg / m s
=
2
1.254 kg / m / s
= 0.653 i.e. 65.3%
Alternatively
Since K α G*2,
4
We write G* = kK ½
4
( )
kK1/2 op.
4
% flooding =
( )
kK1/2 flood
4
( )
K op. 0.32
% flooding = 4 = = 0.653
( )
K flood 0.75
4
Chapter 12 Packed Columns 81
(f) Dimensions of Column
G' 1.056kg / s Mass flow rate
=
Area of column = =
2 Gas flux
G* 0.819kg / m s
Area = 1.289 m2
( )
( )
2
4 1.289m
Diameter of column = =1.281
π
Use stainless steel sheets
since it is bigger than a standard ss pipe size
SO + H O, acidic & corrosive Why ceramic?
2 2
Chapter 12 Packed Columns 82
(g) Packing size to column dia ratio
1281mm
Ratio = = 33.7 √ ok
38mm
Satisfactory, a larger packing size could be used.
Consider 2” intalox, but recalculation needed.
D
>15
is recommended
d
p
Column Diameter Packing Size
<0.6m <25mm
0.3-0.9m 28-38mm
>0.9m 50-75mm
Chapter 12 Packed Columns 83
(h) Height of column
Determine H
toG
(See Coulson and Richardsons Volume 6)
Z = N H
toG toG
Z = (4.74 stages) (1.05 m/stage)
= 4.98 m
(i) ∆P for Irrigated Packing
mm water
∆P = (21 ) (4.98m) = 104.58 mm = 0.105 m water
m packing
From K plot Fig. 11.44
4
∆P = ρgh
= (1000 kg/m3)(9.81 m/s2) (0.105m)
= 1030.0 kg/ms2
Chapter 12 Packed Columns 84
Review slide
Z = H N
toG toG
G dy
y
Z = ∫ 1
use this if K C > K P where K and K are
K aP y y − y L T G T L G
2 the mass transfer
G T e
coefficients,
Mass transfer resistance lies in the vapour phase. C and P are the
T T
concentration and
Z = N H partial pressure of
toL toL
the component and
L x dx a = the effective
Z = ∫ 1
use this if K C < K P wetted area.
K aC x x − x L T G T
2
L T e
Mass transfer resistance lies in the liquid phase
See H & M 1 notes or Pong notes for more details
Chapter 12 Packed Columns 85
(j) ∆P for Dry Packing – 1m - Ergun’s Eq.
∆P 150( 1−ε)2 µu 1.75ρgu2 ( 1−ε)
= c + c
Ζ
ε3
d
2 ε3
d
p p
( )
6 1−ε
For 1 ½’ Intalox saddles, ceramic
d =
p
S
B ε = 0.76 S = 194 m2/m3
B
ρ = 1.177 kg/m3 µ = 1.85 x 10-5 kg/ms (air)
g
kmole
( )( )
0.0332 0.92 29 kg / kmole
s
u = = 0.588m / s
c ( 3)( 2 )
1.177 kg / m 1.281m
d = 7.42 × 10−3 m
p
Z =1m
∆P = 56.35 kg / ms2
Chapter 12 Packed Columns 86
Chapter 12 Packed Columns 87
(k) Total ∆P across tower
∆P = ∆P+ ∆P + ∆P
t i d a
Subscripts t : total
i : irrigated packing (done)
d : dry packing (done)
a : accessories, distributor, gas contraction and expansion,
exit pipe etc.
∆P = 5% of (∆P + ∆P )
a i d
∆P + ∆P = (1030.0 + 56.35) kg/ms2
i d
= 1086.4 kg/ms2
∆P = 1.05 (1086.4 kg/ms2)
t
= 1140.7 kg/ms2
Chapter 12 Packed Columns 88
(l) Wetting Rate
.
(23.39 kg/s)
Vol. liquid rate = = L = 0.024 m3/s
(1000 kg/m3)
(23.39 kg/s)
Vol. rate/column area = = L = 0.0183 m3/m2s
(1.28 m2) (1000 kg/m3)
L 0.0183 m3/m2s
Wetting rate = = = 1.08 x 10-4 m2/s
F 170 m2/m3
P
This is > 2 x 10-5 m2/s √ ok
Wetting Rate (continued)
( )
23.39 kg / s
L* = =18.27 kg / m2 s √ Ok
2
1.28 m
Norman (1961) Recommended:
L* > 2.7 kg/m2s
Chapter 12 Packed Columns 89
Chapter 12 Packed Columns 90
Recommended Wetting Rate
1. Morris (1953)
2
m
> 2 x 10-5 for rings dia. 25 – 75 mm
s
2
m
> 3.3 x 10-5 larger packings
s
2. Norman (1961)
L* > 2.7 kg/m2s
3. Fig 4.3, C&R, vol 2, a nomograph
Chapter 12 Packed Columns 91
(m) Power of Fan
Mass flow rate
(1.056 kg/s)
Vol. gas rate = = 0.897 m3/s
(1.177 kg/m3)
Density
kg
Power = (1140.7 ) (0.897 m3/s) = ∆P V
ms2
kg m2
= 1023.2 = 1.02 kW
s3
Fan eff. = 0.6 let’s say
1.02 kW
Actual Power = = 1.70 kW
0.6
Chapter 12 Packed Columns 92
(n) Inlet & Outlet Liquid Nozzles
(23.39 kg/s)
Vol. liquid flow rate =
(1000 kg/m3)
.
V = 0.0234 m3/s
OD = 141.3 mm
Try DN 125 nozzles, Sch. 40S
ID = 128.2 mm
Thick = 6.55 mm
π (0.128)2
Area = = 0.0129m2
4
.
V
Velocity = = 1.81 m/s √ DN125 is ok
A
Chapter 12 Packed Columns 93
(0) Inlet and Outlet Gas Nozzles
.
Vol. gas rate = V = 0.897 m2/s
OD = 273.1 mm
Try DN 250 nozzles, Sch. 40S
ID = 254.56 mm
Thick = 9.27 mm
( )2
π 0.254
Area = = 0.0507m2
4
V
Velocity = =17.7m / s √ DN 250 is ok
A
Chapter 12 Packed Columns 94
Summary
Internal diameter = 1.3 m
Packing = 1 ½’ intalox saddles ceramics
Height of irrigated packing = 5.0 m
Stainless steel gas distributor and cross beam
Stainless Steel shell – SO + H 0 can be corrosive (implications)
2 2
Fan / Blower 1.7 kW +
One liquid distributor (do we need redistributor?)
Dry packing = 1m height, above liquid distributor (for what?)
∆P = 1.14 kPa
t
% flooding = 65.3%
Hold down grid
Inlet and outlet nozzles
DN 125 Liquid
DN 250 Gas
Chapter 12 Packed Columns 95
Gas out
5.0
Equipment
sketch
Normally we don’t have much
liquid hold up here, and the
Chapter 12 Packed Columns 96
liquid goes to a stripper.
Allow 1 m at top for gas out & disengagement
Allow 2 m at bottom for liquid reservoir and gas in
Max unsupported Packing depth – 5m, near the limit.
Redistributor for liquid not needed – but near limit.
depth 5.0m
= = 3.8
diameter 1.3m
Chapter 12 Packed Columns 97
Other Considerations
External structural supports pressure vessel?
Flanged end cap connections
Man-hole or inspection ports
Instrumentations
If higher velocities are used i.e. operating at higher % of flooding
Then calculations will show that smaller column diameter
Larger ∆P
Larger pumps are needed
End effect:
A lower capital cost
A higher operating cost
Chapter 12 Packed Columns 98
