Gas Absorption & Stripping in Chemical Engineering (Part 3/4)

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1. Introduction to Packed Columns
2. Equipment in Packed Columns
 Packings
 Column Internals
3. Design & Model
 Min. Liquid Ratio
 Column Diameter
 Pressure Drop
 Column Height
 Method 1: HETP – Height Equivalent to a Theoretical Plate
 Method 2: MTU – Mass Transfer Units

 A packed column can be used in absorption as well
 Gas-Liquid interactions are increased here
 Tortuosity increase path length
 Height/Diameter are considered here
 Pressure is to be lost as well
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 Packed tower is known as:
 continuous differential contact equipment
 No specific “stage”
 It is different from the stage-wise
absorption column
 Different Math/Design Approach

 Packings are generally more expensive than plates.
 The difference in column height is not usually
significant if the flow rates are such that efficiencies
are near maximum.
 As a rule of thumb:
 plates are always used in columns of large diameters
 If towers that have more than 20 to 30 stages.

 Conditions favouring packed columns:
 small-diameter columns (less than 0.6m)
 more choices in materials of construction for packings especially
in corrosive service (e.g. plastic, ceramic, metal alloys)
 lower pressure drop (important in vacuum distillation)
 less liquid entrainment · low liquid hold-up, especially suitable
for thermally sensitive material
 foaming liquids can be handled more readily (less agitation of
liquid by the vapour)

 Conditions favouring plate columns:
 variable liquid and/or vapour loads
 low liquid rates
 large number of stages and/or diameter
 high liquid residence time
 dirty service (plate columns are easier to clean)
 presence of thermal or mechanical stress due to large temperature
changes which might lead to cracked packings
 exotherms requiring cooling coils inside column

a) Equilibrium data
 Gas-Liquid
b) Gas and liquid flow rates
c) Solute concentration in two points
d) Individual and overall volumetric mass transfer coefficients
 design of a packed absorption tower.

 (I) Selection of solvent
 (II) Selection of packing
 (III) Calculation of minimum solvent flow rate as well as actual solvent flow rate
 (IV) Column diameter
 (V) Height of column
 (VI) Design of solvent distributors and redistributors
 (VII) Design of gas distributor, packing support, shell, nozzles, column support

So far…

 Various types of packings made of different types of materials of construction are
available:
 Random packings
 Structured packings
 Grid Packaging
Step (II)

 Chemically inert: to the fluids; no reaction, corrosion-resistance
 Mechanical Strength: strong but without excessive weight
 Void Volume: contain adequate passages for both streams without excessive liquid
hold-up or pressure drop
 Surface Area: provide good contact between the liquid and the gas.
 parameter 'a', having the unit m2/m3 or ft2/ft3
 Cost: As low as possible, replaceable and readily available
 Fouling resistance: Packing materials should not trap suspended solids present in
liquid.
 Bigger packing materials generally give low fouling resistance.
Step (II)

 Common materials:
 Metal
 packings are usually preferred because of their superior strength and good
wettability.
 Plastic
 Inexpensive and have sufficient strength
 Poor wettability particularly at low liquid rates.
 Ceramic
 Useful for resisting corrosion at elevated temperatures, where plastic may
not be suitable.
 Good wettability, but inferior strength than metal packings.
Step (II)

 Random packings, as the name implied, are dumped into a column during
installation and allowed to fall in random.
 Small packings poured randomly into a vessel is certainly the more popular and
commonly employed form of packed-tower design.
Step (II)

Step (II)

 When low pressure drop and very high flowrates:
 stacked or oriented packings are used
 Favors decrease in pressure
 Decreases path-length of flow
Step (II)

 Structured packings are considerably more expensive per unit volume than random
packings.
 They come with different sizes and are neatly stacked in the column.
 Structure packings usually offer:
 less pressure drop
 higher efficiency
 Higher capacity
Step (II)

 In dry packing application, the packings are allowed to drop into the column via
the
 (a) chute-and-sock method,
 (b) rope-and-bucket method.
 Dry packing avoids:
 high hydrostatic liquid head
 prevents the introduction of water into a dry process.
 Quicker
 less expensive than wet packing
 Minimises rusting of metal packings.
 Not suitable for plastic packings, as plastic typically floats on water.
Step (II)

 Wet packing applications are preferred when the packings are constructed of breakage-prone
materials
 Ceramic
 Carbon.
 The column is first filled with water and the packings are gently poured down the column.
 The water will:
 cushion the fall of packings
 promotes randomness of settling.
 This tends to increase column capacity
 Improve the column pressure drop characteristics.
 Minimises compression and mechanical damage to packing materials.
 The main disadvantage:
 Need to remove the water after loading and dry the packings.
Step (II)

So far…
Step (VI-VII)

 Liquid Distributor
 Liquid Re-distributor
 Vapour Distributor
 Mist Eliminator (Demister)
 Vortex Breaker
 Impingement baffle / plate
 Packing support
 Retaining devices
Step (VI-VII)

 Its function is to:
 distribute liquid evenly across the tray or packings (e.g. at the reflux inlet)
 Some degree of maldistribution is unavoidable
 As liquid can only be divided into a limited number of streams.
 Liquid distributors are generally classified into:
 pressure distributors (e.g. ladder pipe, spray)
 gravity distributors (e.g. weir type and orifice type).
Step (VI-VII)

 These are used whenever an intermediate liquid feed
needs to be introduced into:
 a packed column
 between packed sections (of tall columns)
 pump-around return to a tray column,
 wherever liquid re-distribution is required.
 Examples include the orifice re-distributors and weir
re-distributors.
 The device also helps to distribute vapour flow in the
column.
 They help to mix and equalise the liquid and vapour
composition throughout the column cross section.
Step (VI-VII)

 These devices are required for:
 introduction of vapour feed from the bottom of the
column
 for the reboiler vapour return to the column.
 A sparger pipe or chimney tray or vapour distributor
with support may be used.
 Some act as support for the packed bed as well.
Step (VI-VII)

 This device is located at the top of the column before the vapour leaves.
 It removes liquid droplets entrained in the vapour stream before the stream leaves
the top of the column.
Step (VI-VII)

Step (VI-VII)

 Avoids vortex formation
 Re-distributes liquid flow to bottom
Step (VI-VII)

 Main Function is to:
 physically support the packed bed
 prevent the downward migration of packing
pieces.
 Packing support should have:
 sufficient open area to permit unrestricted flow
of liquid and vapour.
 Grid supports and corrugated supports are
commonly used.
 A layer of structured packing can sometimes be
used as support.
Step (VI-VII)

 Hold-down plates are used with:
 ceramic or carbon random packing
 rests directly on the packing.
 Prevents fluidization of the packings and restrict
packing movement, which may break the packing
particles.
 Bed limiter is used with:
 metal or plastic packings
 In order to avoid crushing the metal or compressing
the plastic.
 Bed limiters do not rest on packing
 instead they are secured to the column wall by
support ring or bolting clips.
Step (VI-VII)
Bed limiter

 Divided into:
 Top
 Middle
 Bottom
Step (VI-VII)

 Impingement Baffle
 Will not allow feed/reflux flow to spill
 False Downcomer
 Similar to Impingement Baffle
Step (VI-VII)

 Vapour outlets are simpler than liquid outlets.
 Avoiding the presence of entrained liquid
droplets in the vapour streams leaving the
columns.
 Mist eliminators can be installed above the top
tray
 Alternative  a separate knock-out drum can be
used
 Vapour outlet lines should be sloped (i.e. self-
draining) either back to the column or into a
downstream vessel.
Step (VI-VII)

 Feed Nozzles
 Downcomer trapouts
 Chimnay Trays
Step (VI-VII)

 Feed Nozzle
 Used to feed directly flow rates
 Downcomer trapouts
 Used for partial liquid draw-off from tray columns.
 Sufficient residence time for vapour disengagement
 The venting process must be completed downstream of the
column outlet.
 Trapout must be sealed to prevent vapour from flowing up
the downcomer,
Step (VI-VII)

 Chimney tray
 Used as liquid collector or vapour distributor.
 The advantage of chimney trays is that they provide
greater time for vapour disengagement,
 Disadvantage:
 relatively high pressure drop.
 but consume more column height.
Step (VI-VII)

 Bottom Arrangement
 Bottom Outlet (liquid product)

 Bottom feed and reboiler return inlets
 should never be submerged below the liquid level.
 Inlets below the liquid level can also be responsible for excessive
entrainment and premature flooding.
 The section of column wall directly opposite a bottom inlet nozzles
often prone to corrosion and erosion attacks
 especially for cases involving high inlet velocities
 small column diameter
 corrosive chemicals.
 These effects can be reduced by installing an impingement plate.
 The inlet should also not impinge on:
 bottom seal pan
 seal pan overflow
 bottom downcomer.
Step (VI-VII)

 3 arrangements are commonly used for column bottom:
 Unbaffled arrangement
 where both bottom product and reboiler liquid are withdrawn from
a common bottom sump.
 Baffled arrangement
 where the space at the bottom of the column is divided into a
bottom draw-off sump and a reboiler feed sump by a preferential
baffle.
 Once-through reboiler arrangement
 where the reboiler liquid is withdrawn from the bottom downcomer
or from a chimney tray located above the bottom sump.
Step (VI-VII)

 Unbaffled arrangement:
 Advantage of simplicity and low cost.
 They are preferred in small column (less than 3-ft diameter) where baffles
are difficult to inspect and maintain.
 They are used for kettle reboilers because the bottom product is
withdrawn from the reboiler surge compartment, not from the column
bottom sump.
 They are also used with forced-circulation reboilers as the large circulation
rate makes it difficult to achieve a steady liquid overflow across the
baffle.
 Because it is difficult to deflect liquid raining from a packed bed
above the reboiler sump into the reboiler side of the preferential
baffle, unbaffled arrangement is normally used in packed column.
Step (VI-VII)

 Baffled or once-through arrangements:
 Usually preferred with thermosiphon reboilers in large columns (more
than 3-ft diameter).
 They can supply:
 a constant liquid head to the reboilers
 maximise bottom sump residence time when vapour disengagement is
the main consideration.
Step (VI-VII)

 Bottom Outlet
 Achieve the required phase separation between vapour and liquid
 Achieve the required surge capacity.
 Designed for avoiding the presence of vapour in the liquid outlet
 bottom sumps
 chimney trays
 surge drums
 On the other hand:
 downcomer trapouts  usually designed to allow for the presence of vapour.
Step (VI-VII)

 Presence of vapour in liquid outlet lines can cause:
 Pump cavitation; erosion; column instability; etc…
 Some of the common causes are:
 Insufficient residence time for vapour disengagement from the liquid.
 Liquid arriving the pan or sump from which it is withdrawn almost always contains entrained vapour
bubbles.
 Frothing (waterfall pool effect) is caused by impact of falling liquid on liquid surface in the sump
or draw pan.
 Vortexing occurs because of intensification of swirling motion as liquid converges towards an
outlet.
 Vortexing promotes entrainment of vapour into the draw-off line.
 Vortex breakers can be installed to prevent vortexing.
Step (VI-VII)

So far…

 Main Scope of the Design:
 Flow Rate
 Minimum Solvent Flow Rate
 (seen in Tray Columns)
 Column diameter
 Analysis of pressure drop across the packed bed.
 Packed height
 HETP method.
 Method of Transfer Units.

So far…
Step (IV)

 In determining the column diameter:
 The limiting (maximum) gas velocity must be defined
 The higher the gas velocity:
 the greater the resistance that will be encountered by the down-flowing liquid
 the higher the pressure drop across the packings.
 Too high gas velocity:
 Leads to flooding  the liquid fills up all the column
 High pressure  will crush and damage the packings in the column
Step (IV)

 We will begin our analysis by examining the relationship
between:
 Gas pressure drop
 Gas velocity
 The horizontal axis is the logarithmic value of the gas
velocity G
 The vertical axis is the logarithmic value of pressure
drop per height of packing
 Due to the fluid friction created by the flow of gas and
liquid around the individual solid packing materials
Step (IV)

 Assume: a dry packing (i.e. no liquid flow, L = 0)
 pressure drop increases as gas velocity increases
 This according to the linear relationship as shown by line a-a.
 This is a straight line on a log-log plot.
Step (IV)

 Assume: liquid flowing in the column
 the packings now become wetted (irrigated).
 Part of void volume in the packings now filled with liquid
 This reduces the cross-sectional area available for gas flow.
 At the same gas velocity:
 the pressure drop is higher for wetted packings compared to
dry packings.
 For L = 0 vs. L = 5.
 The line for dP/L under wetted condition lies to the left of line a-a.
Step (IV)

 For a constant liquid flow (say L = 5000), at low to moderate
gas velocity G
 the pressure drop characteristics is similar to that of dry
packings
 i.e. section b-c of the plot is still straight on log-log plot.
 There is:
 orderly trickling of the liquid down the packings
 no observable liquid being trapped among the packings
 no liquid hold-up
Step (IV)

 As the gas velocity is increased further
 the pressure drop increased.
 some liquid started to be retained in the packings.
 When point c is reached
 the quantity of liquid retained in the packed bed increases
significantly.
 There is a change in slope of the line at point c
 Pressure drop increases more rapidly with G.
 Point c is known as the loading point
 the liquid starts to accumulate (load) in the packings.
Step (IV)

 From point c to d to e:
 there is a sharp increase in pressure drop at higher G
 there is a greater amount of liquid hold-up
 a gradual filling of the packing voids with liquid (bottom to top)
 the column is slowly "drowned" in the liquid.
 At point e
 there is another sharp change in the slope.
 At this point the entire column is filled liquid
 the gas now has to bubble through the liquid in the packing voids.
 The gas pressure drop is now very high.
 Point e is known as the flooding point.
 The gas velocity at this point is known as the flooding velocity
(limiting velocity).
Step (IV)

 Points to note :
 at constant liquid rate
  gas pressure drop increases with gas velocity.
 at higher liquid rate
  the loading and flooding points occur at lower gas pressure
drop.
 at constant gas velocity
  the gas pressure drop is higher at larger liquid rate.
 each liquid rate has its own loading and flooding points.
Step (IV)

 Operation of a gas absorption column is not practical
above the loading point.
 For optimum design  the recommended gas velocity
 1/2 of the flooding velocity.
 Also, design can be based on:
 pressure drop condition
 below the pressure drop at which flooding would occur.
Step (IV)

 Summary:
 For optimum design  the recommended gas velocity
 1/2 of the flooding velocity.
 Avoid loading
 Avoid at any cost flooding

 Most packed columns consist of cylindrical vertical vessels.
 The column diameter is determined so as to safely avoid flooding and operate in the
preloading region with a pressure drop of no greater than 1.2 kPdm of packed height
 equivalent to 1.5 in. of water head per foot of packed height
 Operating below loading and flooding is important
 Flooding Velocity is required!
 Flow parameter (X)
 superficial gas velocity (VFG)

 Flooding data for packed columns with countercurrent flow of gas and liquid were first
correlated successfully by Sherwood et al. (1938) in terms of the flow parameter
 X =flow paramater
 L’ = mass flow rate of liquid (kg/s)
 V’ = mass flow rate of gas (kg/s)
 ρG = density of gas (kg/m3)
 ρL = density of liquid (kg/m3)

 The superficial gas velocity at flooding, VGF, as embedded in the dimensionless term
 Where,
 g = gravity
 VFG = superficial gas velocity at flooding
 The ratio a/ε3 is a function of the packing only
 a = specific surface area of packing, m2/m3
 ε = porosity or void fraction; dimensionless
 Fp = packing factor ft-1 or ft2/ft3

 Leva ( 1954) used experimental data on ring and
saddle packings to extend the Shenvood et al. (1938)
 Based on flooding correlation and included lines of
constant pressure
 This chart is known as the generalized pressure drop
correlation (GPDC).
GPDC

GPDC
Y: pressure-drop parameter defined in equation (4-6); dimensionless.

 The flooding curve can be accurately described by the polynomial regression
 Recall:
 pressure-drop parameter defined in equation (4-6); dimensionless.

 It has been found that the pressure drop at flooding is strongly dependent on the
packing factor for both random and structured packings.
 Kister and Gill (1991) developed the empirical expression
 Where
 dPflood = Drop in pressure, Pa/m
 Fp has units of ft2/ft3.
DPflood = 93.9Fp
0.7

 Packed columns are designed based on either of two criteria
1. The fractional approach to flooding gas velocity
2. The maximum allowable gas-pressure drop.

 1. Use Equation to get VFG 
 X = based on the flow rates
 L, G, and densities are typically given/can be calculated
 Y = From the graph/equation
 Based on the packing and Cs
 Solving conveniently
VG =
Cs
rG
rL - rG
é
ë
ê
ù
û
ú
0.5
Cs =
Y
FpmL
0.1
æ
è
ç
ö
ø
÷
0.5
VG =
Y(rL - rG )
rGFpmL
0.1
é
ë
ê
ù
û
ú
0.5
VG =
Y(rL - rG )
rGFpmL
0.1
é
ë
ê
ù
û
ú
0.5
Y = e
- 3.5021+1.28ln X+0.11093(ln X)2é
ë
ù
û
VG =
e
- 3.5021+1.28ln X+0.11093(ln X)2é
ë
ù
û
(rL - rG )
rGFpmL
0.1
é
ë
ê
ê
ù
û
ú
ú
0.5

 2. Select Flooding criteria f= (0.5-0.7)
 No more than 80%
 3. Use the following Diameter Correlation 
 Where,
 D = diameter, m
 Qg = volumetric flow rate of Gas, m3/s
 f = flooding factor (fractional approach)
 VGF = The superficial gas velocity at flooding
 PI = 3.1416

 Absorbers and strippers are usually designed for gas pressure drops:
 200 to 400 Pdm of packed depth
 Use the GPDC chart for estimation  Estimate Diameter
 This is the specified maximum pressure drop

 Alternatively, use a correlation:
 Billet and Schultes (1991a) correlation
 KW = wall factor in Billet-Schultes pressure-drop correlations; dimensionless
 Yo = dry-packing resistance coefficient in Billet-Schultes pressure-drop correlations;
dimensionless.
 The ratio a/ε3 is a function of the packing only
 a = specific surface area of packing, m2/m3
 ε = porosity or void fraction; dimensionless
DPo
Z
= Yo
a
e3
rGvG
2
2
1
KW

 KW  Wall factor in Billet-Schultes pressure-drop correlations; dimensionless.
 dp  particle size; m.
1
KW
=1+
2
3
1
1- e
æ
èç
ö
ø÷
dp
D
dp = 6
1- e
a
æ
èç
ö
ø÷

 The dry-packing resistance coefficient (a modified friction factor), Yo, is given by
the empirical expression
 ReG  Reynolds Number for Gas flow
 Cp  a packing constant determined from experimental data and tabulated for a
 number of packings
Yo = Cp
64
ReG
+
1.8
ReG
0.08
æ
èç
ö
ø÷
ReG =
vGdprGKW
(1- e)mG

 GAS:
 Air containing 5 mol% NH
 at a total flow rate of 20 kmol/h
 a packed column operating at 293 K and 1 atm
 90% of the ammonia is scrubbed by a counter
current flow
 LIQ:
 1500 kg/h of pure liquid water.
Benitez, Mass Transfer Operations.
Example 4.3 Pressure Drop in Beds Packed with First- and Third-Generation Random Packings

 For 25-mm ceramic Raschig rings
 (ai) Estimate the superficial gas velocity at flooding
 (aii) Estimate pressure drop at flooding
 (bi) Estimate the column inside diameter at 70% of flooding
 (bii) Pressure drop for operation at 70% of flooding

 Get X, flow parameter…
 Get G’
X=
L'
G'
æ
èç
ö
ø÷
rG
rL
æ
èç
ö
ø÷
0.5
G' = MWavgxG
MWavg = x1MW1 + x2MW2 = 0.95x29 + 0.05x17 = 28.4
G' = MWavgxG = 28.4
kg
kmol
æ
èç
ö
ø÷ 20
kmol
h
æ
èç
ö
ø÷
1h
3600s
æ
èç
ö
ø÷ = 0.1577kg/s

 Get L’
 Since dilute, assume properties of water
 Get Densities of G’ and L’ (average…)
 For gas:
 For liquid...
L' = 1500
kg
h
æ
èç
ö
ø÷
1h
3600s
æ
èç
ö
ø÷ = 0.416kg / s
ravg-gas =
PMWavg
RT
=
1atm( ) 28.4g / mol( )
8.314
J
molK
æ
èç
ö
ø÷ 293K( )
=1.18kg / m3
ravg-liq =1000kg / m3

 Get X
X =
0.416kg / s
0.1577kg/s
æ
èç
ö
ø÷
ravg-gas
ravg-liq
æ
è
ç
ö
ø
÷
0.5
=
0.416kg / s
0.1577kg/s
æ
èç
ö
ø÷
1.18
1000
æ
èç
ö
ø÷
0.5
X = 0.0909
X=
L'
G'
æ
èç
ö
ø÷
rG
rL
æ
èç
ö
ø÷
0.5
=
0.416kg / s
0.1577kg/s
æ
èç
ö
ø÷
ravg-gas
ravg-liq
æ
è
ç
ö
ø
÷
0.5

 Get Y… either graphically or via equation
lnYflood = - 3.5021+1.028ln X + 0.11093(ln X)2
éë ùû
lnYflood = - 3.5021+1.028ln(0.0909)+ 0.11093(ln(0.0909))2
éë ùû
lnYflood = - 3.5021+ -2.4651+ 0.63788[ ]
lnYflood = -1.675
Yflood = e-1.675
= 0.1873

 Now, for… For 25-mm ceramic Raschig rings
Fp = 179 ft2
/ft3
a = 190m2
/ m3
e = 0.68
Ch = 0.577
Cp = 1.329

 Get Cs,flood conditions:
Yflood = FpCsflood
2
mL
0.1
mL = 1cp = 0.001kg / ms
Csflood =
Yflood
FpmL
0.1
=
0.1873
179x(0.0010.1
)
Csflood = 0.04569m/s

 Get Vg:
Csflood = vG , flood
rG
rL - rG
é
ë
ê
ù
û
ú
0.5
vG , flood
=
Csflood
rG
rL - rG
é
ë
ê
ù
û
ú
0.5 =
0.04569m/s
1.18
1000 -1.18
æ
èç
ö
ø÷
0.5
vG , flood
= 1.3298m/s
This is (ai)

 Get dP flood:
 NOTE: Fp = ft2/ft3 and yields Pa/m
DPflood = 93.9(Fp)0.7
= 93.9(179)0.7
DPflood = 3,545.3Pa/m This is (aii)

 At 70% conditions…
 Calcualte expected diameter
 Get Volumetric Flow Rate of Gas 
f = 0.70
QG =
G'
ravg
=
0.1577kg / s
1.18kg / m3
= 0.13364m3
/s
D=
4QG
fVGp
é
ë
ê
ù
û
ú
0.5
=
4(0.13364m3
/s)
0.7x(1.3298m/s)(3.1416)
é
ë
ê
ù
û
ú
0.5
D = 0.4275m
D=0.43m This is (bi)

 Calcualte expected pressure Drop according to Billet et al.
 Get all data requirements…
f = 0.70
This is (bi)
DPo
Z
= Yo
a
e3
rGvG
2
2
1
KW
Fp = 179 ft2
/ft3
a = 190m2
/ m3
e = 0.68
Ch = 0.577
Cp = 1.329

 Get dp, then Kw, then Re, then Yo
 For particle dp
 For Kw
This is (bi)
d =
6(1- e)
a
=
6(1- 0.68)
190
= 0.010105m
1
KW
= 1+
2
3
1
1- e
æ
èç
ö
ø÷
dp
D
1
KW
= 1+
2
3
1
1- 0.68
æ
èç
ö
ø÷
0.0101m
0.43m
= 1.048934
KW =
1
1.048934
= 0.9532

 For ReG
 For Yo
This is (bi)
ReG =
vGdprGKW
(1- e)mG
ReG =
(0.9303m / s)(0.010105m)(1.18kg/m3
)(0.9532)
(1- 0.68)1.85x10-5
kg / m is
ReG =1796.3
Yo = Cp
64
ReG
+
1.8
ReG
0.08
æ
èç
ö
ø÷
Yo = 1.329m / s
64
1796.3
+
1.8
1796.30.08
æ
èç
ö
ø÷ = 1.3608m/s

 Now, substitute all data:
This is (bii)
Po
Z
= Yo
a
e3
rGvG
2
2
1
KW
Po
Z
= 1.3608m/s
190
(0.683
)
1.18kg / m3
( )(0.93032
)
2
1
0.9532
Po
Z
= 440.48
Pa
m

 Repeat Ex. 14…
 GAS:
 Air containing 5 mol% NH
 at a total flow rate of 20 kmol/h
 a packed column operating at 293 K and 1 atm
 90% of the ammonia is scrubbed by a counter
current flow
 LIQ:
 1500 kg/h of pure liquid water.
Benitez, Mass Transfer Operations.
Example 4.3 Pressure Drop in Beds Packed with First- and Third-Generation Random Packings

 For 25-mm metal Hi-flow rings
 (ci) Estimate the superficial gas velocity at flooding
 (cii) Estimate pressure drop at flooding
 (di) Estimate the column inside diameter at 70% of flooding
 (dii) Pressure drop for operation at 70% of flooding

 Get X, flow parameter…
 Get G’
X=
L'
G'
æ
èç
ö
ø÷
rG
rL
æ
èç
ö
ø÷
0.5
G' = MWavgxG
MWavg = x1MW1 + x2MW2 = 0.95x29 + 0.05x17 = 28.4
G' = MWavgxG = 28.4
kg
kmol
æ
èç
ö
ø÷ 20
kmol
h
æ
èç
ö
ø÷
1h
3600s
æ
èç
ö
ø÷ = 0.1577kg/s

 Note that:
 X and Y are independent of the packing…
 What else changes?
 Why?
 What can we do to avoid over-work

 Now… For 25-mm metal Hi-flow rings
a = 202.9
Fp = 42
e = 0.962
Ch = 0.799
Cp = 0.689

 Get Cs,flood conditions:
Yflood = FpCsflood
2
mL
0.1
mL = 1cp = 0.001kg / ms
Csflood =
Yflood
FpmL
0.1
=
0.1873
179x(0.0010.1
)
42
0.09432

 Get Vg:
rG
rL - rG
é
ë
ê
ù
û
ú
0.5
vG , flood
=
Csflood
rG
rL - rG
é
ë
ê
ù
û
ú
0.5 =
0.04569m/s
1.18
1000 -1.18
æ
èç
ö
ø÷
0.5
vG , flood
= 1.3298m/s
This is (ci)
0.09432
2.744

 Get dP flood:
 NOTE: Fp = ft2/ft3 and yields Pa/m
= 93.9(179)0.7
DPflood = 3,545.3Pa/m This is (cii)
1,285.10
42

 Calcualte expected diameter
 Get Volumetric Flow Rate of Gas 
f = 0.70
QG =
G'
ravg
=
0.1577kg / s
1.18kg / m3
= 0.13364m3
/s
D=
4QG
fVGp
é
ë
ê
ù
û
ú
0.5
=
4(0.13364m3
/s)
0.7x(1.3298m/s)(3.1416)
é
ë
ê
ù
û
ú
0.5
D = 0.4275m
D=0.43m
This is (di)
2.744
0.2976m

 Calcualte expected pressure Drop according to Billet et al.
 Get all data requirements…
f = 0.70
DPo
Z
= Yo
a
e3
rGvG
2
2
1
KW
Fp = 179 ft2
/ft3
a = 190m2
/ m3
e = 0.68
Ch = 0.577
Cp = 1.329
a = 202.9
Fp = 42
e = 0.962
Ch = 0.799
Cp = 0.689

 Get dp, then Kw, then Re, then Yo
 For particle dp
 For Kw
d =
6(1- e)
a
=
6(1- 0.68)
190
= 0.010105m
1
KW
= 1+
2
3
1
1- e
æ
èç
ö
ø÷
dp
D
1
KW
= 1+
2
3
1
1- 0.68
æ
èç
ö
ø÷
0.0101m
0.43m
= 1.048934
KW =
1
1.048934
= 0.9532
0.962
202.2
0.001127
0.001127
0.2976m
0.962
1.0664
0.9377

 For ReG
 For Yo
ReG =
vGdprGKW
(1- e)mG
ReG =
(0.9303m / s)(0.010105m)(1.18kg/m3
)(0.9532)
(1- 0.68)1.85x10-5
kg / m is
ReG =1796.3
Yo = Cp
64
ReG
+
1.8
ReG
0.08
æ
èç
ö
ø÷
Yo = 1.329m / s
64
1796.3
+
1.8
1796.30.08
æ
èç
ö
ø÷ = 1.3608m/s
1.9208
0.001
0.9377
0.962
3,325.56
0.689
3,325.56 3,325.56
0.6615m/s

 Now, substitute all data:
This is (dii)
Po
Z
= Yo
a
e3
rGvG
2
2
1
KW
Po
Z
= 1.3608m/s
190
(0.683
)
1.18kg / m3
( )(0.93032
)
2
1
0.9532
Po
Z
= 440.48
Pa
m
0.6615m/s
0.962
202.2
1.92m/s
0.9377
181.49 Pa/m

 When molasses is fermented to produce a liquor containing ethanol, a C02-rich vapor
containing a small amount of ethanol is evolved.
 The alcohol will be recovered by countercurrent absorption with water in a packed-bed
tower. The required recovery of the alcohol is 97%.
 GAS:
 Rate= 180 kmol/h, at 303 K and 110 kPa.
 The molar composition of the gas is 98% CO, and 2% ethanol.
 LIQUID:
 Pure liquid water at 303 K will enter the tower at the rate of 151.5 kmol/h
 This is 50% above the minimum rate required for the specified recovery
 TOWER:
 Packed with 50-mm metal Hiflow rings
 maximum pressure drop of 300 P/m of packed height.
Benitez, Mass Transfer Operaitons. Example 4.4 Design of a
Packed-Bed Ethanol Absorber

 (a) Determine the column diameter for the design conditions.
 (b) Estimate the fractional approach to flooding conditions
Benitez, Mass Transfer Operaitons. Example 4.4 Design of a
Packed-Bed Ethanol Absorber

 (A) Column Diameter
 Get X, based on L’, G’ and densities
 Get Y, based on equation of X
 Get Data for the Packing
 Get the superficial velocity
 Calculate Diameter given the correlation.
Seen in
Ex. 14!

 Get X
Seen in
Ex. 14!
L' = MWavgL = 18
kg
kmol
æ
èç
ö
ø÷ 151.5
kmol
h
æ
èç
ö
ø÷
1h
3600s
æ
èç
ö
ø÷ = 0.7575
kg
s
G' = MWavgG = 44
kg
kmol
æ
èç
ö
ø÷ 180
kmol
h
æ
èç
ö
ø÷
1h
3600s
æ
èç
ö
ø÷ = 2.2
kg
s
rliq = 1000
kg
m3
rgas =
PMWav
RT
=
(101.320kPa)(44g / mol)
8.314
m3
Pa
K - mol
æ
èç
ö
ø÷ (303K)
= 1.769
kg
m3
X=
L'
G'
æ
èç
ö
ø÷
rG
rL
æ
èç
ö
ø÷
0.5
=
0.7575
kg
s
2.2
kg
s
1.769
kg
m3
1000
kg
m3
æ
è
ç
ç
ç
ö
ø
÷
÷
÷
0.5
= 0.01453

 Get Y
Seen in
Ex. 14!
lnYflood = - 3.5021+1.028ln X + 0.11093(ln X)2
éë ùû
lnYflood = - 3.5021+1.028ln(0.015)+ 0.11093(ln(0.015))2
éë ùû
lnYflood = - 3.5021+ -4.3172 +1.9565[ ]
lnYflood = -1.1414
Yflood = e-1.1414
= 0.31937

 Now… For 50-mm metal Hi-flow rings
a = 92.3
Fp = 16
e = 0.977
Ch = 0.876
Cp = 0.421

 Get Vg flooding at given conditions
Seen in
Ex. 14!
Yflood = FpCsflood
2
mL
0.1
mL = 0.631cp = 0.000631kg / ms
Csflood =
Yflood
FpmL
0.1
=
0.319
16x(0.0006310.1
)

 Get Vg flooding at given conditions
Seen in
Ex. 14!
rG
rL - rG
é
ë
ê
ù
û
ú
0.5
vG , flood
=
Csflood
rG
rL - rG
é
ë
ê
ù
û
ú
0.5 =
0.2041m/s
1.769
1000 -1.769
æ
èç
ö
ø÷
0.5
vG , flood
= 4.848m/s

 Get dP
 We need max. 300 Pa/m
 Iteration begins!
 Assume a f = 0.5
Seen in
Ex. 14!
= 93.9(16)0.7
DPflood = 653.95Pa/m

 Calculate the volumetric flow rate… (the same for all iterations)
 Now... Get Diameter given f = 0.5
Seen in
Ex. 14!
QG =
G'
ravg
=
2.2kg / s
1.769kg / m3
=1.2436m3
/s
D=
4QG
fVGp
é
ë
ê
ù
û
ú
0.5
=
4(1.2436m3
/s)
0.7x(4.484m/s)(3.1416)
é
ë
ê
ù
û
ú
0.5
Df =0.5 =12.29m

 Repeat all other iterations
Seen in
Ex. 14!
D=
4QG
fVGp
é
ë
ê
ù
û
ú
0.5
=
4(1.2436m3
/s)
fx(4.484m/s)(3.1416)
é
ë
ê
ù
û
ú
0.5
Df =0.5 = 0.8402m
Df =0.6 = 0.7670m
Df =0.7 = 0.7101m
Df =0.8 = 0.6642m

 Calcúlate Pressure Drops… recall that dp= 300 Pa/m max
Seen in
Ex. 14!
Df =0.5 = 0.8402m
Df =0.6 = 0.7670m
Df =0.7 = 0.7101m
Df =0.8 = 0.6642m
DPo
Z
= Yo
a
e3
rGvG
2
2
1
KW

 Only dp is constant, regardless of “f”
Seen in
Ex. 14!
d =
6(1- e)
a
=
6(1- 0.977)
190
= 0.0002491mCheck out Full COURSE:

 For KW, D changes
Seen in
Ex. 14!
Df =0.5 = 0.8402m
Df =0.6 = 0.7670m
Df =0.7 = 0.7101m
Df =0.8 = 0.6642m
1
KW
= 1+
2
3
1
1- e
æ
èç
ö
ø÷
dp
Df
1
KW
= 1+
2
3
1
1- 0.977
æ
èç
ö
ø÷
0.0002491m
Df
1
KW
= 1+
0.007217
Df
1.0085
1.0094
1.0101
1.0110

 Assume a
Seen in
Ex. 14!
ReG =
fvGdprGKW
(1- e)mG
ReG =
f (4.484m / s)(0.00024m)(1.769kg/m3
)
(1- 0.977) 1.45x10-5
kg / m is( )(1.001)
ReG = 5,708.3f
ReGf =0.5 = (5,708.3)0.5 = 2,854
ReGf =0.6 = (5,708.3)0.6 = 3,425
ReGf =0.7 = (5,708.3)0.7 = 3,995
ReGf =0.7 = (5,708.3)0.8 = 4,566

 Get Yo
Seen in
Ex. 14!
Yo = Cp
64
ReG
+
1.8
ReG
0.08
æ
èç
ö
ø÷
Yo = 0.421m / s
64
ReG
+
1.8
ReG
0.08
æ
èç
ö
ø÷
Yo f =0.5 = 0.421m / s
64
2854
+
1.8
2854( )0.08
æ
è
ç
ö
ø
÷ = 0.4104
Yo f =0.6 = 0.421m / s
64
3425
+
1.8
3425( )0.08
æ
è
ç
ö
ø
÷ = 0.4030
Yo f =0.7 = 0.421m / s
64
3995
+
1.8
3995( )0.08
æ
è
ç
ö
ø
÷ = 0.397
Yo f =0.8 = 0.421m / s
64
4566
+
1.8
4566( )0.08
æ
è
ç
ö
ø
÷ = 0.3920

 Assume a
Seen in
Ex. 14!
Po
Z
= Yo
a
e3
rGvG
2
2
1
KW
Po
Z
= 0.40
92.3
0.9773
1.77( fxvG )2
2
1
1
Po
Z
= 35.234(f ×4.484)2
Po
Z
= 708.34f2
Po
Z f =0.5
= 708.34(0.5)2
= 177.085Pa/m
Po
Z f =0.6
= 708.34(0.6)2
= 255.0Pa/m
Po
Z f =0.7
= 708.34(0.7)2
= 347.086Pa/m
Po
Z f =0.8
= 708.34(0.8)2
= 453.33Pa/m

 Assume linear interpolation will not yield extreme error:
Seen in
Ex. 14!
Po
Z f =0.6
= 708.34(0.6)2
= 255.0Pa/m
Po
Z f =0.65
= 708.34(0.6)2
= 299.27Pa/m
Df =0.5 = 0.8402m
Df =0.6 = 0.7670m
Df =0.7 = 0.7101m
Df =0.8 = 0.6642m
D =
0.7670 + 0.7101( )m
2
= 0.73855
D=0.74m

So far…
Step (V)

 There are two main methods to approach this:
 Height Equivalent to a Theoretical Plate (HETP)
 Method of Transfer Units (MTU)
Step (V)

 When packings are used instead of trays:
 the same enrichment of the vapour will occur over a certain height of packings
 this height is termed as the height equivalent to a theoretical plate (HETP).
 The composition of solute from entry to exit of the packed tower is represented by
operating line
 every point indicates some location in the packed tower
 On the other hand, in tray tower, few points (number of trays) in the operating line
represents the conditions in the trays.
Step (V)

 As all sections of the packings are physically the same:
 it is assumed that one equilibrium (theoretical) plate is
represented by a given height of packings.
 The required height of packings for any desired
separation is given by:
 Htotal = ( HETP ) x ( No. of ideal trays required )
Step (V)

 The ratio between packing height to number of trays required for the same
separation:
 Height equivalent to theoretical plate (HETP).
Solving
Step (V)

 HETP depends:
 size and type of packing
 flow rate of gas and liquid
 concentration of solute
 physical and transport properties as well as
equilibrium relationship
 uniformity of liquid and gas distribution.
Step (V)

 Empirical correlations are restricted to limited applications.
 The main difficulty:
 failure to account for the fundamentally different action of tray vs packed columns.
 In industrial practice, the HETP concept is used to convert empirically:
 the number of theoretical trays to packing height.
 HETP is used to characterize packing
 A good packing has small HETP.
Step (V)

Step (V)

 For an absorption process:
 Tray required  8
 Total height of the column is 5.2 m
 Diameter  3 m
 The column was assigned to a process engineer. After several experinetation, he
setted a packing sized approx. 3 meters.
 (a) Calculate the HETP calculated

 (a) Calculate the HETP calculated
Z = HETPxN
HETP =
Z
N
HETP =
3m
8m
HETP = 0.375

 Assume we are using the following packing:
 Total height used  5.8m
 Flow Number  2.5
 (a) Select betst Packing (EX) vs. (DX)
 (b) Get HETP
 (c) Get expected tray number given EMGE = 0.65

 (a) Best Design
 Given the flow number, only Dx can operate
 Select Dx
 (b) HETP given the flow number.
 Between 0.06 and 0.10
 Select 0.08
HETP =
HETP
m
æ
èç
ö
ø÷ x Z( )= (0.08)(5.8m) = 0.464m/tray

 (c)
Nideal =
Z
HETP
=
5.8m
0.464m/tray
= 12.5
Nreal =
Nideal
EMGE
=
12.5
0.65
=19.23
Nreal » 20

 IMPORTANT NOTE!
 This method had been largely replaced by the Method of Transfer Units.
Step (V)

 This method is more appropriate because:
 the changes in compositions of the liquid and vapour phases occur differentially in a
packed column
 Compared to the stepwise fashion as in trayed column.
 Height of packings required can be evaluated via
 gas-phase or the liquid-phase.
 The packed height (z) is calculated using the following formula:
 z = N x H
 where
 N = number of transfer units (NTU) – dimensionless
 H = height of transfer units (HTU) - dimension of length
Step (V)

HTU NTU
Step (V)
NONINTEGRAL: quantity is called ‘height if
transfer units’ (HTU) and designated as HtG.
The INTEGRAL part of Equation (4.14) is called
number of gas phase transfer units as NtG.

 Process can be applied to either:
 Gas Balance
 Liquid Balance
Step (V)

 The number of transfer units (NTU) required
 is a measure of the difficulty of the separation.
 A single transfer unit gives the change of composition of one of the
phases equal to the average driving force producing the change.
 The NTU is similar to the number of theoretical trays required for
trayed column.
 NTU = No. Trays
 A larger number of transfer units will be required for a very high
purity product.
Step (V)
HTU NTU

 The height of a transfer unit (HTU) is a measure of
the separation effectiveness of the particular packings for a
particular separation process.
 It incorporates the mass transfer coefficient
 The more efficient the mass transfer(i.e. larger mass transfer
coefficient) 
 the smaller the value of HTU.
 Larger mass transfer coefficient leads to the smaller value of HTU.
 The values of HTU can be estimated from empirical correlations
or pilot plant tests, but the applications are rather restricted.
Step (V)
HTU NTU

 Recall that it is hard to get the local concentrations, interphase “i” values
 Overall concentrations in equilibrium “*” are preferred
 We will use “O” subscript
 Process can be applied to either:
 Gas Balance; “G” subscript
 Liquid Balance “L” subscript
Step (V)

 For the gas-phase
 we have: z = NOG x HOG
 KY is the overall gas-phase mass transfer coefficient.
 "a" is the packing parameter
 y, mol fraction in 1,2
Step (V)

 "a" is the packing parameter that we had seen earlier (recall the topic on column
pressure drop, e.g. Table 6.3) that characterize the wetting characteristics of the
packing material (area/volume).
 Normally, packing manufacturers report their data with both KY and "a" combined as
a single parameter.
 Since KY has a unit of mole/(area.time.driving force), and "a" has a unit of
(area/volume), the combined parameter KY a will have the unit of
mole/(volume.time.driving force), such as kg-mole/(m3.s.mole fraction).
 As seen earlier, other than mole fraction, driving force can be expressed in partial
pressure (kPa, psi, mm-Hg), wt%, etc.
Step (V)

 y1* is the mole fraction of solute in vapour that is in equilibrium with the liquid of mole
fraction x1 and y2* is mole fraction of solute in vapour that is in equilibrium with the liquid of
mole fraction x2 .
 The values of y1* and y2* can be obtained from the equilibrium line as previously covered
Step (V)

 [ Point P (x, y) as shown is any point in the column. The concentration difference
driving force for mass transfer in the gas phase at point P is (y - y*) as shown
previously, this time no subscripts are shown. ]
 NOTE: Both equilibrium line and operating line are straight lines
under dilute conditions.
 Alternatively, equilibrium values y1* and y2* can also be calculated using Henry's
Law ( y = m x, where m is the gradient) which is used to represents the equilibrium
relationship at dilute conditions.
 Thus, we have: y1* = m x1 ; y2* = m x2
Step (V)

 Similarly for the liquid-phase we have:
 z = NOL x HOL
 KX is the overall liquid-phase mass transfer coefficient
 "a" is the packing parameter seen earlier.
 Again, normally both KX and "a" combined as a single parameter.
Step (V)

 Likewise, x1* is the mole fraction of solute in liquid that is in equilibrium with the
vapour of mole fraction y1 and x2* is mole fraction of solute in liquid that is in
equilibrium with the vapour of mole fraction y2 .
 Alternatively, x1* = y1 /m and x2* = y2 /m.
Step (V)

 (x1* - x1) is the concentration difference driving force for mass transfer in the liquid
phase at point 1 (bottom of column) and (x2* - x2) is the concentration difference
driving force for mass transfer in the liquid phase at point 2 (top of column).
Step (V)

 Using either gas-phase or liquid-phase formula should yield the same required
packing height :
Step (V)

 Changing “LOCAL” to “OVERALL” mass transfer coefficients…
 The developed equations:
 Inputs:
 NOG:
 y1*,y2* (from x1 and x2 respectively)
 HOG:
 G
 KY
 A
 y1, y1* (from x1)

 Steps:
1. Get Equilibrium line (Henry’s, etc…)
2. Get x1, y1 (Bottom Column Conditions)
3. Get x2, y2 (Top Column Conditions)
4. Draw Operation Line
5. Get y1* and y2*
 y1*  from x1, in the equilbirium (horizontal)
 y2*  from x2, in the equilbirium (horizontal)
6. Calculate using equations:
1. NOG
2. HOG
Step (V)

 MTU = Method of Transfer Units
 The Number of Transfer Units (NTU) and Height of
Transfer Units (HTU) such as NOG, HOG
 should not be confused with the number of theoretical
trays (N), and the height equivalent to theoretical
plate (HETP) respectively.
Step (V)
Htotal = (No. of ideal trays required ) x (HETP)

 When the operating line and equilibrium line are straight and parallel:
 NTU = N ; and HTU = HETP
 Otherwise, the NTU can be greater than or less than N
Step (V)

 When the operating line is straight but not parallel, we have the following
relationships:
 A = L/(mG); the absorption Factor
Step (V)

 Consider the ethanol absorber of Ex. 16…
 packed with 50-mm metal Hiflow rings.
 The system ethanol-C02-water at the given
temperature and pressure obeys Henry’s law
 m = 0.57
 (a) Estimate the packed height required to recover
97% of the alcohol
 using pure water at a rate 50% above the minimum
 And gas-pressure drop of 300 Pa/m

Benitez, Mass Tranfer Operaitons, 2nd.
Example 5.4 Packed Height of an Ethanol absorber
Ky = 0.075407kmol
s-m3

 Since Ky is given, use overall conditions, for gas…
 For packed height:
 Get NOG first  does not needs MTC
 Then calcualte HOG  Requires MTC

 The amount of ethanol absorbed required:
 This will be used in y2
nethanol-removed = (180kmol / h) 0.02
kmol ×eth
kmol ×mix
æ
èç
ö
ø÷ 97%( )
nethanol-removed = 3.5kmol / h.
nethanol-left = (180kmol / h) 0.02
kmol ×eth
kmol ×mix
æ
èç
ö
ø÷ 3%( ) = 0.108kmol/h

 Calculate
 Does note requires MTC
 Get y2 
NtOG
y2 =
mol×ethanol
G
=
0.108kmol / h
180kmol / h
= 0.0006
A =
L
mV
A =
151.5kmol / h
(0.57)(180kmol / h)
A = 1.4766
Get A 
NtOG =
ln
y1 - mx2
y2 - mx2
1-
1
A
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
1-
1
A

 Calculate
 Does note requires MTC
NtOG
NtOG =
ln
y1 - mx2
y2 - mx2
1-
1
A
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
1-
1
A
NtOG =
ln
0.02 - 0.57(0)
0.0006 - 0.57(0)
1-
1
1.4766
æ
èç
ö
ø÷ +
1
1.4766
é
ë
ê
ù
û
ú
1-
1
1.4766
NtOG =
ln(11.4372)
0.32276
NtOG = 7.552

 Get HOG
a = p
D
2
æ
èç
ö
ø÷
2
= p
0.74m
2
æ
èç
ö
ø÷
2
= 0.86
HtOG =
G
Kyah
HtOG =
G
Kyah
=
180kmol / h( )
1h
3600s
æ
èç
ö
ø÷
Kyah
=
HtOG =
0.05kmol/s
Ky 0.86
=
0.05kmol/s
0.075407kmol
s-m3( )x(0.86m2
)
= 0.77108m

 Get packed height
Z = HOG NOG
Z = (0.77108m)(7.35m)
Z = 5.67m

 Solute A is to be absorbed from a binary mixture: (7.5% of A) in a packed tower.
 Based on flooding calculation, a tower diameter of 1.2 m is selected.
 Total gas flow rate is 60 kmol/h.
 The exit gas must not contain more than 0.2% of solute A.
 Solute free liquid B enters from the top of the tower at 40 kmol/h.
 The gas phase and liquid phase mass transfer coefficients based on mole ratio unit are:
 The equilibrium line Equation is Y=0.63X.
 Specific interfacial area of gas-liquid contact (ā) is 71 m2 /m3 .
NPTEL – Mass transfer Web - Lec3
kx = 2.05kmol
m2
h
ky =1.75kmol
m2
h

 (a) Calculate packing height required for the desired separation.

 (a) Calculate packing height required for the desired separation.
Z = NtOG HtOG
NtOG =
ln
y1 - mx2
y2 - mx2
1-
1
A
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
1-
1
A
HtOG =
G
Kyah

 Get A, and y1,y2, x1,x2
 Since 7.5%... Not that dilute… still assume dilute!
 From Mass Balance:
L2
x2
G2
y2
G1
y1
L1
x1
G = 60kmol / h
L = 50kmol / h
y1 = 0.075
y2 = 0.002
G1y1 + L2 x2 = G2 y2 + L1x1
60(0.075)+ 40(0) = (60)(0.002)+ 40(x1)
x1 =
60(0.075)- (60)(0.002)
40
= 0.1095
x1 = 0.1095

 Calculate NOG
 Get A factor
NtOG =
ln
y1 - mx2
y2 - mx2
1-
1
A
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
1-
1
A
A =
L
mG
=
40
(0.63)(60)
A = 1.0582
L2
x2
G2
y2
G1
y1
L1
x1

L2
x2
G2
y2
G1
y1
L1
x1
NtOG =
ln
0.075 - (0.63)0
0.002 - (0.63)x0
1-
1
1.058
æ
èç
ö
ø÷ +
1
1.058
é
ë
ê
ù
û
ú
1-
1
1.058
NtOG =
ln(2.132)
0.05482
NtOG = 13.764

 Calculate KY since it is required
1
Ky
=
1
ky
+
m
kx
1
Ky
=
1
1.75
+
0.63
2.05
1
Ky
= 0.8787
Ky = 1.138kmol
m2
h-Dy
L2
x2
G2
y2
G1
y1
L1
x1

 Calculate HOG
HtOG =
G
Kyah
=
60
kmol
h
æ
èç
ö
ø÷
(1.138kmol
m2
h-Dy
) 71
m2
m3
æ
èç
ö
ø÷
HtOG = 0.74259m
L2
x2
G2
y2
G1
y1
L1
x1

 Calculate Z
L2
x2
G2
y2
G1
y1
L1
x1
Z = NtOG HtOG
Z = (13.764)(0.74259m)
Z=10.1824m
Z= »10.2m

 From Exercise#20…
 If we wanted to remove 99.5% instead of 0.02% specification...
 (a) How much higher must the column be?
L2
x2
G2
y2
G1
y1
L1
x1

 Get A, and y1,y2, x1,x2
 Since 7.5%... Not that dilute… still assume dilute!
 From Mass Balance:
L2
x2
G2
y2
G1
y1
L1
x1
G = 60kmol / h
L = 40kmol / h
y1 = 0.075
y2 = ?
nAleft = (60kmol / h)(0.075)(0.005) = 0.0225kmol/h
nAabsorbed = (60kmol / h)(0.075)(0.995) = 4.4775kmol/h
y2 =
nAleft
G
=
0.0225kmol/h
60kmol / h
y2 = 0.000375
y2 previous
y2 new
=
0.002
0.000375
= 5.33

L2
x2
G2
y2
G1
y1
L1
x1
NtOG =
ln
0.075 - (0.63)0
0.000375 - (0.63)x0
1-
1
1.058
æ
èç
ö
ø÷ +
1
1.058
é
ë
ê
ù
û
ú
1-
1
1.058
NtOG =
ln(12.26)
0.05482
NtOG = 45.20

 Calculate KY since it is required
1
Ky
=
1
ky
+
m
kx
1
Ky
=
1
1.75
+
0.63
2.05
1
Ky
= 0.8787
Ky = 1.138kmol
m2
h-Dy
L2
x2
G2
y2
G1
y1
L1
x1

 Calculate Z
L2
x2
G2
y2
G1
y1
L1
x1
Z = NtOG HtOG
Z = (45.20)(0.74259m)
Z=33.44
Z= » 33.4m
Zratio =
33.4
10.2
= 3.27x

 Acetone in air is being absorbed by water in a packed tower having a cross-sectional
area of 0.186 m2 at 293 K and 1 atm.
 At these conditions, the equilibrium relation is given by y=1.186x
 The inlet air contains 2.6 mole % acetone and the outlet 0.5 %.
 The air flow is 13.65 kmol/hr
 Pure water inlet flow is 43.56 kmol/hr.
 Film coefficients for the given flow in the tower:
 kya = 3.8×10-2 kmol/s·m 3
 kxa = 6.2×10-2 kmol/s·m 3 .
 (a) Determine the tower height.

 The equation:
 Get NOG:
Z = NOGHOG
NtOG =
ln
y1 - mx2
y2 - mx2
1-
1
A
æ
èç
ö
ø÷ +
1
A
é
ë
ê
ù
û
ú
1-
1
A
A =
L
mG
=
43.56
(1.186)(13.65)
A = 2.69

NtOG =
ln
0.026 -1.186(0)
0.005 -1.186(0)
1-
1
2.690
æ
èç
ö
ø÷ +
1
2.690
é
ë
ê
ù
û
ú
1-
1
2.690
NtOG =
ln(3.638)
0.6282
NtOG = 2.055

 Get HOG
 For which, we will need Ky 1
Ky
=
1
ky
+
m
kx
1
Kya
=
1
kya
+
m
kxa
1
Kya
=
1
0.038
+
1.186
0.062
Kya = 0.022

 Get HOG
HOG =
G
Kya
=
G'
Kya× Ac
HOG =
(13.65kmol / h)
1h
3600s
æ
èç
ö
ø÷
0.022kmol / m( )(1.86m2
)
HOG = 0.9264m

 Get N
Z = HOG NOG = (0.9264m)(2.0348)
Z =1.90m Check out Full COURSE:

Gas Absorption & Stripping in Chemical Engineering (Part 3/4)

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