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CHAPTER 4: GAS ABSORPTION
CHAPTER / CONTENT Definition, Application and Notation Used in Gas Absorption Conditions of Equilibrium Between Liquid and Gas According to Raoult’s Law The Mechanism of Absorption Packed – tower → Description and Design
Plate tower → Description and Design
Definition, Application and Notation Used in Gas Absorption A type of mass transfer operation (separation) between gas and liquid system. The removal of one or more selected components from a mixture of gases. Or A soluble vapor is absorbed from its mixture with an inert gas by means of a liquid the soluble gas is more soluble. Common example of gas absorption: Ammonia – Air System Ammonia can be absorbed by passing the gases (NH3 – air) into water where ammonia will be dissolved in water. The ammonia (solute) can then be recovered by distillation and the absorbing liquid can be either discarded or reused. Acetone – Air mixture
Conditions of Equilibrium Between Liquid and Gas According to Raoult’s Law The two phases (gas and liquid) when brought into contact tend to reach equilibrium.
Consider air – water system, the water in contact with air evaporates until the air is saturated with water vapor, and the air is absorbed by the water until it becomes saturated with the individual gases. In any mixture of gases, the degree to which each gas is absorbed is determined by its partial pressure at a given temperature and pressure. When a single gas (solute) and a liquid (solvent) are brought into contact (until equilibrium), the resulting concentration of dissolved gas (solvent) in liquid is called gas solubility (at T and P). At fixed temperature, solubility concentration increased when pressure increased .
Conditions of Equilibrium Between Liquid and Gas According to Raoult’s Law Partial pressure of solute in gas phase (kN/m2) 1.3 6.7 13.3 26.7 66.7
Concentration of solute in water kg/1000 kg water Ammonia
Sulfur dioxide
Oxygen
11 50 93 160 315
1.9 6.8 12 24.4 56
0.08 0.13 0.33
Ammonia – Very soluble Sulfur Dioxide – A moderate soluble Oxygen – Slightly soluble
Conditions of Equilibrium Between Liquid and Gas According to Raoult’s Law In any mixture of gases the solubility of each gas depends on partial pressure Solubility also depends on Temperature (Solubility ↑ as T ↑). Recall Raoult’s Law for ideal solution:
p A x A PAo p A partial pressureof A in gas phase x A mol fraction of A in liquid phase PAo vapor pressureA at that temperature
In G.A.; *
*
P P x where P Equilibriu m partial pressure o
Conditions of Equilibrium Between Liquid and Gas According to Raoult’s Law Equilibrium Distribution (Solubility Curve) In G.A, feed is a gas and enters at bottom of column and the solvent is fed at top column, the absorbed gas and solvent leave out the bottom and unabsorbed components leave as gas from the top. Liquid / solvent is well below its boiling point and gas molecules are diffusing into liquid.
y
Equilibrium curve
(y,x)
x Figure 1 – Equilibrium curve
Conditions of Equilibrium Between Liquid and Gas According to Raoult’s Law Film concept / Theory in mass transfer Due to this concept: Gas molecules must diffuse from main body of the gas phase to the gas – liquid interphase, then Cross this interface into the liquid side and finally diffuse from the interface into the main body of the liquid. For dilute concentration of most gases, and over a wide range for some gases, equilibrium relationship is given by Henry’s Law.
pA H CA p A partial pressureof A in gas phase C A concentration of componentin liquid H Henry's constant
The Mechanism of Absorption
THE TWO – FILM THEORY
RATE OF ABSORPTION
EVALUATION OF MASS TRANSFER COEFFICIENT
The Two – Film Theory Developed by Whitman.
MAIN BULK OF GAS
GAS FILM BOUNDARY
PARTIAL PRESSURE (P) OF SOLUBLE GAS
Pi
LIQUID FILM B D
GAS FILM
MAIN BULK OF LIQUID
Ci E
CL
MOLAR CONCENTRATION OF SOLUTE IN A LIQUID
A
PG
LIQUID FILM BOUNDARY
According to this theory, material is transferred in the bulk of the phase by convection currents, and concentration differences are considered as negligible.
Figure 2 – Concentration profile for absorbed component A On either side of this interface it suppose that the currents die out and that there exists a thin film of fluid flowing through which the transfer occurs only to molecular diffusion.
The Two – Film Theory According to Fick’s Law, the rate transfer by diffusion is proportional to the concentration gradient and to the area of interface over which the diffusion is occurring. The direction of transfer of material across the interface, is, however, not dependent on the concentration difference, but on the equilibrium relationship. The controlling factor – the rate of diffusion through the two film. From Figure 2: PG - Partial pressure in the bulk of the gas phase Pi - Partial pressure at interface
CL - Concentration in the bulk of the liquid phase Ci - Concentration at interface
The Two – Film Theory According to the two – film theory: The concentration at the interface are in equilibrium The resistance to transfer is centred in the thin films on either side of the wall Assumptions of two – film theory: Steady state – concentration at any position do not change with time Interface between the gas and liquid phase is a sharp boundary Laminar film – exist at the interface on both sides of the interface Equation exist at the interface No chemical reaction (rate of diff. across gas phase = rate of diff. across liquid phase)
MAIN BULK OF GAS
GAS FILM BOUNDARY
PARTIAL PRESSURE (P) OF SOLUBLE GAS
Pi
LIQUID FILM B D
GAS FILM
MAIN BULK OF LIQUID
Ci E
Figure 2 – Concentration profile for absorbed component A
CL
MOLAR CONCENTRATION OF SOLUTE IN A LIQUID
A
PG
LIQUID FILM BOUNDARY
The Two – Film Theory
Rate of Absorption The process of absorption may be regarded as the diffusion of a soluble gas A into a liquid. The molecules of A have to diffuse through a stagnant gas film and then through a stagnant liquid film before entering the main bulk of liquid. The rate of absorption of A per unit time over unit area is given: N A' kG PA1 PA2
kG - Gas film transfercoefficient
The rate of diffusion in liquids is much slower than in gases, and mixtures of liquids may take a long time to reach equilibrium. For the rate of absorption of A into liquid: N A' k L C A1 C A2
kL
- liquid film transfercoefficient
Rate of Absorption In steady state process of absorption, the rate of transfer of material through the gas film will be the same as that through the liquid film. The general equation for mass transfer can be written as:
N A' kG PG Pi k L Ci CL PG - Partial pressure in the bulk of the gas phase
Pi - Partial pressure at interface CL - Concentration in the bulk of the liquid phase Ci - Concentration at interface N’A - Overall rate of mass transfer (mol/unit area.time) Therefore:
kG Ci CL k L PG Pi
Rate of Absorption This condition can be shown graphically as below where ABF is equilibrium curve.
D
A
Figure 3 – Driving forces in the gas and liquid phase
PG – Pi
PG
E’ Ci – CL
Pi PB
B
F CL
Ci
Ce
Point D ( CL , PG ) – represents conditions in bulk of gas and liquid Point A ( Ce, PG ) – represents concentration Ce in the liquid in equilibrium with PG in the gas. Point B ( Ci, Pi ) – represents concentration Ci in the liquid in equilibrium with Pi in the gas, and gives conditions at the interface. Point F (CL, Pe ) – represents the partial pressure Pe in the gas in equilibrium with CL in the liquid.
Rate of Absorption Then the driving force causing transfer in the gas phase:
PG Pi DE The driving force causing transfer in the liquid phase:
Ci CL BE Then,
PG Pi k L Ci CL kG
The concentration at the interface (point B) are found by drawing a line through D of slope – kL/kG to cut the equilibrium curve in B.
Rate of Absorption Overall Coefficients To obtain a direct measurement of the values kL and kG, we require the measurement of the concentration at the interface. Need to define two overall coefficient, KL and KG N A' KG PG Pe K L Ce CL
K G - Overallgas phasecoefficient K L - Overallliquid phase coefficient
The rate of transfer A can now be written as:
N A' kG PG Pi k L Ci CL KG PG Pe K L Ce CL
Eq.1
Rate of Absorption kG PG Pi K G PG Pe 1 1 PG Pe K G kG PG Pi 1 1 PG Pi 1 K G kG PG Pi kG
From Eq. (1):
Pi Pe PG Pi
kG PG Pi k L Ci CL 1 1 P P G i kG k L Ci CL
Insert equation above into Eq. (2): 1 1 PG Pi 1 Pi Pe K G kG PG Pi kG PG Pi 1 1 1 P P P P G i i e K G kG k L Ci C L PG Pi 1 1 1 P P i e K G k G k L Ci C L
Eq.3
Eq.2
Rate of Absorption where term Pi Pe = average slope of equilibrium curve and, Ci C L
if the solution obeys Henry’s Law (H): H
dP Pi Pe dC Ci CL
Therefore Eq. (3) becomes: 1 1 1 P P i e K G kG k L Ci CL 1 1 H K G kG k L
Similarly: 1 1 1 K L k L HkG
and
1 H KG K L
Eq.3
Rate of Absorption Factors Influencing the Mass Transfer coefficient Very soluble gas (e.g NH3 in H2O). - resistance is so small where H ≈ negligible. So kG ≈ KG • Low solubility gas (e.g O2 in H2O) - resistance is in the liquid H is large. So kL ≈ KL • Moderately soluble gas - both film offer resistance, so kG ≠ KG and kL ≠ KL Example: Show that from rate of transfer of A: a)
1 1 H K G kG k L
b)
1 1 1 K L k L HkG
Evaluation of Mass Transfer Coefficient Mass transfer coefficient α Diffusivity (D) α 1/film thickness (zg) for gas For gas:
kG
Dg
For liquid:
zg
kL
DL zL
Problem arises when measuring kG, kL since we do not know the value of film thickness. However, one piece of equipment where the surface area is known as the: Wetted – Wall column D Slow flowing film of water - laminar flow condition
H Sometimes turbulent flow could occur
GAS
Glass tube
Evaluation of Mass Transfer Coefficient Surface area of film D H Flux, N KG P P* N unit in mol/(m2.s)
P* = Equilibrium pressure
Method: Measure gas concentration entering Measure gas concentration leaving Surface area of film is its area of the tube Calculate the concentration driving force at inlet and outlet and take log mean difference KG
NA P P*
Evaluation of Mass Transfer Coefficient Example The data given below were obtained from a wetted – wall column with a constant liquid flow rate. Molar Gas Flow rate, G (kmol/s)
Overall Mass Transfer Coefficient, KG (kmol/s.m2 (kN/m2)
0.03 0.06 0.12 0.18
157.8 210.6 261.0 285.6
kG also related to the gas flow rate by:
kG AG
0.82
where A is constant
For molar gas flow rate of G = 0.1 kmol/s, evaluate the individual mass transfer coefficient (kG and kL) and overall mass transfer coefficient (KG) if H = 20 (kN/m2)/kmol
Evaluation of Mass Transfer Coefficient Solution From Rate of Absorption:
1 1 H K G kG k L Assuming kL constant and given, kG AG
0.82
where A is constant
1 1 H K G AG 0.82 k L
Eq.a
Equation (a) is straight line equation (y=mx+c) where: y → 1 KG
m →
1A
0.82 x → 1G
c →
H kL
Evaluation of Mass Transfer Coefficient Solution Construct graph: y – axis → 1 K G 0.82 x – axis → 1 G
G (kmol/s)
KG x106 (kmol/s.m2 (kN/m2))
1 G 0.82
0.03
157.8
17.732
6.337 x 10-3
0.06
210.6
10.044
4.748 x 10-3
0.12
261.0
5.689
3.831 x 10-3
0.18
285.6
4.080
3.501 x 10-3
1 KG
Evaluation of Mass Transfer Coefficient Solution 7.0E-03 6.0E-03
1/KG x 10 -6
5.0E-03 4.0E-03
y = 2.079E-04x + 2.653E-03
3.0E-03 2.0E-03 1.0E-03 0.0E+00 0
2
4
6
8
10
12
14
16
18
20
1/G0.82
From the graph, straight line obtained: y = 2.079x10-4x + 2.653 x 10-3
m = 1 A = 2.079x10-4 c = H kL
= 2.653 x 10-3
Evaluation of Mass Transfer Coefficient Solution For molar gas flow rate of G = 0.1 kmol/s, evaluate the individual mass transfer coefficient (kG) and overall mass transfer coefficient (KG) if H = 20 (kN/m2)/kmol G (kmol/s)
KG x106 (kmol/s.m2 (kN/m2))
1 G 0.82
0.10
248.3
6.607
1 K G x 10-6 4.027 x 10-3
From equation obtained: y = 2.079x10-4x + 2.653 x 10-3 y 2.079 10 4 x 2.653 10 3 y 2.079 10 4 (6.607) 2.653 10 3 y 4.027 10 3 1 4.027 10 3 KG K G 248.3
Evaluation of Mass Transfer Coefficient Solution From equation, slope: 1 2.079 10 4 A 1 A 4810 4 2.079 10
m
Individual mass transfer coefficient for gas, kG: kG AG
0.82
kG 48100.10
0.82
kG 728.0
Evaluation of Mass Transfer Coefficient Solution From equation, y – intercept c
H 2.653 10 3 kL
Individual mass transfer coefficient for liquid, kL:
For H 20 kN/m 2 /kmol H 20 kL c 2.653 10 3 k L 7538.6
Packed – tower → Description and Design
INTRODUCTION TO PACKED – TOWER
PRESSURE DROP AND FLOODING IN DETERMINATION OF TOWER DIAMETER
DETERMINATION OF HEIGHT OF TOWER
PACKED TOWER –
Introduction to Packed – Tower Packed towers are used for continuous counter – current in absorption. The tower in Figure 4 consists of a cylindrical column containing: A gas inlet and distributing space at the bottom A liquid inlet and distributing device at the top A gas outlet at the top A liquid outlet at the bottom A packing filling in the tower. A large intimate contact between the liquid and gas is provided by the packing
Figure 4 Packed tower flows
Introduction to Packed – Tower Common types of packing which are dumped at random in the tower are shown in Figure 5. Packing are available in size of 3 mm to about 75 mm and mostly are made of materials such as clay, porcelain, metal or plastic. High void spaces of 65 – 95% are characteristics of good packings. The packings permit relatively large volumes of liquid to pass countercurrent to the gas flow through the openings with relatively low pressure drops for the gas.
Figure 5 Typical random or dumped tower packings (a) Rashig Ring (b) Berl Saddle (c) Pall Ring (d) Intalox Metal (e) Jaeger Metal Tri - Pack
Pressure Drop and Flooding in Packed Towers In a given packed tower with a given type and size of packing and with a definite flow of liquid, there is an upper limit to the rate of gas flow, called the flooding velocity.
The tower cannot be operated at gas flow velocity above flooding velocity. At a low gas velocities, the liquid flows downward through the packing, essentially uninfluenced by the upward gas flow.
As the gas flow rate is increased at low gas velocities, the pressure drop is proportional to the flow rate to the 1.8 power. At a gas flow rate called the loading point, the gas starts to hinder the liquid down flow, and local accumulations or pools of liquid start to appear in the packing (liquid holdup). The pressure drop of the gas starts to rise at a faster rate. As the flow rate of gas increased, the liquid holdup or accumulation increases.
Pressure Drop and Flooding in Packed Towers At the flooding point, the liquid can no longer flow down through the packing and is blown out with the gas. In actual operating tower, the gas velocity is well below flooding velocity. The optimum economic gas velocity is about one – half or more of flooding velocity. Flooding velocity depends on: Type of packing / packing factor Size of packing Liquid mass velocity Limiting pressure drop, Pflood 0.115Fp0.7 where Pflood : Pressuredrop at flooding Fp
: Packing Factor
Pressure Drop and Flooding in Packed Towers
Pressure Drop and Flooding in Packed Towers Example 1 Ammonia is being absorbed in a tower using pure water at 25OC and 1 atm abs pressure. The feed rate is 1440 lbm/h (653.2 kg/h) and contains 3.0 mol% of ammonia in air.
The process design specifies a liquid – to – gas mass flow rate ratio GL / GG of 2/1 and use 1-in. metal Pall rings. Calculate the pressure drop in the packing and the gas mass velocity at flooding. Using 50% of the flooding velocity, calculate the pressure drop, gas and liquid flows, and tower diameter. Repeat (a) above by use Mellapak 250Y structured packing.
Pressure Drop and Flooding in Packed Towers Solution 1 Find the required data to be used in Figure 10.6-5 (Geankoplis pp. 660)
Given mol fraction for ammonia = 0.03 Given mol fraction for air = 0.97
Mwt ammonia = 17 Mwt air = 29
Average molecular weight of entering gas: M av yi x Mwti 0.03 x 17 0.97 x 29 28.64
ρg
PM RT
1 atm 28.64 3
82.06
g mol
cm atm 298.15 K mol.K
1.17110 3 g cm 3 @ 0.0730 lbm/ft 3
Pressure Drop and Flooding in Packed Towers Solution 1 From Appendix A.2-4, the water viscosity = 0.8937 cP. From A.2-3, the water density is 0.99708 g/cm3 1lbm g 106 cm 3 0.3048 m3 ρL 0.99078 3 x x x 61.85 lbm /ft 3 3 3 cm 1m 1 ft 453.59237 g 3
1 Centistokes = 10-2 cm2/s μ 1 cm 3 10- 2 g ν 0.8937cP x x 0.902 x 10- 2 cm 2 s 0.902 centistok es ρ 0.99078 g 1 cP cm.s
From Table 10.6-1, for 1-in, Pall rings, Fp=56 ft-1. Using equation 10.6-1, ΔPflood 0.115Fp0.7 0.115 56
0.7
1.925 in. H2 O / ft packing height
Pressure Drop and Flooding in Packed Towers Solution 1 x-axis for Figure 10.6-5: GL ρG G G ρL
0 .5
2 0.0730 1 61.85
0 .5
0.06871
For flow parameter of 0.06871 (abscissa) and pressure drop 1.925 in/ft at flooding, a capacity parameter (ordinate) of 1.7 is read off the plot. From y – axis:
1.7 υG ρG ρL ρG Fp ν 0.05 0.5
0.5
1.7 υG 0.07310 61.85 0.07310 560.50.9020.05 0.2561 0.5
υG 6.6381 ft s
Pressure Drop and Flooding in Packed Towers
Pressure Drop and Flooding in Packed Towers Solution 1 GG ρG υG 0.07310
lbm ft lbm 6 . 6381 0 . 4852 at flooding ft 3 s ft 2 s
Using 50% of the flooding velocity for design, GG 0.5 0.4852
lbm lbm 0 . 2426 ft 2 s ft 2 s
Liquid flow rate, GL 2 0.2426
lbm lbm 0.4852 2 2 ft s ft s
To calculate the pressure drop at 50% flooding, GG 0.2426 lbm 2
ft s lbm and GL 0.4852 2 . The new capacity parameter is 0.5 x 1.7 = 0.85. ft s
By using value of 0.85 and flow parameter of 0.06871 (abscissa), a value of 18 in H2O/ft is obtained.
Pressure Drop and Flooding in Packed Towers Solution 1 The tower cross sectional area, At Feedrate 1440lb m ft 2 s 1 hr At x 1.6488 ft 2 GG hr 0.2426 lb m 3600 s At
Dt 2
Dt 2
4 4 At
4 1.6488 2.099 ft 2 3.142
Dt 2.099 ft 2
0.5
`1.448 ft
Pressure Drop and Flooding in Packed Towers Example 2 Ammonia (NH3) is being removed from air by scrubbing with water in a packed tower with 6 mm ceramic Berl Saddles (Cf = 900. The gas entering at 1.2 m3/s contains 15% NH3. The water enters at a rate of 4 kg/s and has a specific gravity of 1, viscosity of 2.5 x 10-3 kg/m.s. The gas mixture enters at 27OC and 1 bar (0.987 atm). Given Molecular weight for ammonia = 17, water = 18. Calculate the diameter of the packed tower when 80% of NH3 is removed and the pressure drop is 400 N/m2 per m packing.
Pressure Drop and Flooding in Packed Towers Solution 2
Lin = 4 kg/s xA = 0
Gout = ? yA = ?
27OC 0.987 atm 80% NH3 is removed
Lout = L’ Lwater = 4 kg/s xA = ?
Gin = 1.2 m3/s yA = 0.15
Find the required data to be used. Given mol fraction for ammonia = 0.15 Given mol fraction for air = 0.85
Mwt ammonia = 17 Mwt air = 29
Average molecular weight of entering gas: M av yi x Mwti 0.15 x 17 0.85 x 29 27.2
Pressure Drop and Flooding in Packed Towers Solution 2 ρg
PM RT
0.987 atm 27.2 3
82.06
g mol
cm atm 300.15 K mol.K
1.090 10 3 g cm3
Given Gin = 1.2 m3/s g m 3 10 6 cm3 Mass of G in ρg V 1.090 10 1.2 1308 g/s @ 1.308 kg/s cm3 s 1 m3 3
cm3 0.987 atm 1.2x10 PV s 48.0871 mol/s @ 0.04809 kmol/s Mol of G in , n 3 cm atm RT 82.06 300.15 K mol.K 6
Mass of G NH3 in 0.15 x0.04809
kg kmol x17 0.1226 kg/s s kmol
Pressure Drop and Flooding in Packed Towers Solution 2 Mass of NH 3 absorbedin water 0.80 x 0.1226 kg/s 0.0981 kg/s
Mass of Gout Mass of gas in Mass of NH 3 absorbed 1.308 0.0981 1.2099 kg/s
Mass of Lout Mass of water Mass of NH 3 absorbed 0.0981 4 4.0981 kg
From information, the water viscosity = 2.5 x 10-3 kg/m.s. Water density is 1000 kg/m3. Since the larger flow quantities are at the bottom of absorber, the diameter will be chosen to accommodate the bottom condition. From Treybal: L' ρG abscissa (x-coordinate) = (Refer to bottom condition) 0 .5 G' ρL ρG 0 .5
0.5 kg 1.090 x 10 -3 s 0.10 kg - 3 0.5 1.308 1 1.090 x 10 s
4.0981
G ' 2C f L0.1 J
G L G g C
Pressure Drop and Flooding in Packed Towers
Pressure Drop and Flooding in Packed Towers Solution 2 N m2 At pressure drop of 400 and x-coordinate = 0.10, m
y-coordinate = 0.066 0 .1
G' 2C f μL J
ρG ρL ρG g c
0.066
g 0.066 1.091000 1.091 2 G '' 0.066 G L 0.1G c 0.145 3 0.1 C J 900 2.5 10 1 f L kg G '' 0.38 2 m s kg 1.308 G' s 3.442 m 2 Cross sectional area, A G'' 0.38 kg m 2 .s
πD 2 A 4
D
4A π
4 3.442 π
D 2.09 m
Exercise
A packed tower is to be designed for a counter current contact of an NH3- air mixture with water to wash out NH3 from the gas. The conditions are: Gas in: Gas out: All NH3 is removed. Flow rate = 1.5 m3/s Temperature = 27 °C Pressure = 1 bar (0.987 atm) Contains 8 mol % NH3 Liquid in: Flow rate = 4.8 kg/s Density = 996 kg/m3 Viscosity = 2.5 x 10-3 kg/m.s Packing used is 38mm Raschig ring (Cf = 95, gc = 1) (a) Calculate the flow rate of liquid out. (b) If the pressure drop of the packed tower is 400 N/m2, by using the diagram, calculate the required diameter for the tower.
Determination of Height of Tower Component denoted in Figure 6: Gm = Mols of inert gas / (unit time) (unit cross – section of tower)
Lm = Mols of inert liquid / (unit time) (unit cross – section of tower)
Gm
Lm
y’2
x’2
y’
x‘ dZ
y‘ + dy’
x’ + dx’
Y = Mols of soluble gas A / mol of inert gas B in gas phase X = Mols of soute A / mol of inert solvent C in liquid phase Mass balance over differential / small section of column: N A' AdZ kG aPG Pi Adz
N A' AdZ k L aCi CL Adz
Gm
Lm
y’1
x‘1
Figure 6 Countercurrent absorption tower
wh ere Adz Vol. of small sectionof tower
N’A = mol / (m2.s) a = Interfacial area / Volume of column
Determination of Height of Tower From the tower,
N A' AdZ Gm Ady
kG aPG Pi A dZ Gm Ady
From Dalton and Raoult’s Law,
Therefore,
y
PG PT
yPT PG
yi
Pi PT
yi PT Pi
and
Gm dy kG aPT y yi dZ dZ
Gm dy kG aPT y yi
Integrating both sides; get:
Gm Z kG aPT
y1
y2
dy y yi
Z = height of tower
Determination of Height of Tower Similarly for concentration in liquid phase:
Z
Lm dx k L a xi x
Normally, Z is written in terms of KG.a & KL.a & in terms of mol fraction:
Z
Gm dy K G aPT y y*
Z H OG NOG
Z = height of tower HOG = height of transfer unit – constant. NOG = number of transfer unit – constant
Determination of Height of Tower And for concentration in liquid phase:
Z
Lm K L aCT
dx x* x
Simplifying:
Z H OG NOG
For gas phase based on equilibrium concentration
Z H OL NOL
For liquid phase based on equilibrium concentration
The number of transfer unit (NOG) can be calculated using several method which will be discussed later.
Methods for Evaluation of NOG
Graphical Method
Log Mean Driving Force Method
Colburns Method
Graphical Methods Based on interface concentration:
Z H G NG H L N L Based on Area, A:
Gm y1 y Lm x1 x
y1 y
Lm x1 x Gm
y y Lm 1 Gm x1 x
Operating line equation relates concentration in gas phase to that in liquid phase. From operating line above:
Graphical Methods y L x1 x y1 G
L L x x1 y1 ope ratingline G G L whe re : gradie nt @ slope G L x1 y1 inte rse ction with y - axis G y
y
Operating line
Equilibrium line – normally linear at dilute concentration
x
Graphical Methods Example An acetone – air mixture containing 0.015 mol fraction of acetone will be reduced to 1% of this value by countercurrent absorption in fresh pure water in packed tower. The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s. For the system, Henry’s law holds and y* = 1.75 x where y* is the mol fraction of acetone in the vapor in equilibrium with x mol fraction in liquid. Calculate the height of the tower / absorber if HOG = 0.3 m.
Data given: Cross sectional A of column = 1 m2 Mwt air = 29 Mwt acetone = 58 Mwt H2O = 18
Graphical Methods Solution Gm=1.0 kg/m2.s
Lm=1.6 kg/m2.s x2 = 0
y2 = 0.00015 2
Given:
Reduced to 1% 1 y1 = 0.015 Gm=1.0 kg/m2.s
x1 = ? Lm=1.6 kg/m2.s
Cross sectional A of column = 1 m2 Mwt air = 29 Mwt acetone = 58 Mwt H2O = 18 HOG = 0.3 m y* = 1.75 x
Graphical Methods Solution The given mass flow rate in kg/m2.s. Need to convert into kmol /s. Gm
Gm in mass unit Area M wt
kg 2 m s 1m 2 0.03448 kmol Gm kg s 29 kmol 1.0
Find slope of the graph. L L x x1 y1 ope ratingline G G L 0.08889 kmol / s slope 2.578 G 0.03448 kmol / s y
Lm
Lm in mass unit Area M wt
kg 2 m s 1m 2 0.08889 kmol Lm kg s 18 kmol 1.6
Graphical Methods Solution Calculate the mol ratio for each location. Location 1
Location 2
y1
A 0.015 0.0152 Gm 1 0.015
y2
A 0.00015 0.00015 Gm 1 0.00015
x1
A ? Lm
x2
A 0 0 Lm 1 0
Need to find x1 in order to plot operating line. L L x x1 y1 G G y 2.578 x C y
Graphical Methods Solution Use coordinate at location 2 in order to find C value. y 2.578 x C 0.00015 2.5780 C C 0.00015
Operating line, y 2.578x 0.00015
Insert value at location 1 in order to find x1 value y1 2.578 x1 0.00015 0.015 2.578 x1 0.00015 x1
0.015 0.00015 0.00576 2.578
Construct operating line by plotting coordinate at location 1 and 2
Location 1
Location 2
x1, y1 0.00576 , 0.015
x2 , y2 0 , 0.00015
Graphical Methods
Operating line
Equilibrium line
No. of stages = 11
NOG = 11
Graphical Methods Solution Height of absorber, Z
Z H OG N OG
Z 0.3m 11 Z 3.3m
Log Mean Driving Force Method dy y y*
or
dx x* x
To evaluate NOG, we can take an average driving force in log mean.
N OG
y1 y2 y y* 1 y y* * y y 1 ln y y* 2
N OG
2
y1
(y - y*)1 (y - y*)
y1 y2
ln y y *
y1* Hx1
y - line
y2* Hx2
y2 (y - y*)2
y* - line
Log Mean Driving Force Method Example An acetone – air mixture containing 0.015 mol fraction of acetone will be reduced to 1% of this value by countercurrent absorption in fresh pure water in packed tower. The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s. For the system, Henry’s law holds and y* = 1.75 x where y* is the mol fraction of acetone in the vapor in equilibrium with x mol fraction in liquid. Calculate the height of the tower / absorber if HOG = 0.3 m.
Data given: Cross sectional A of column = 1 m2 Mwt air = 29 Mwt acetone = 58 Mwt H2O = 18
Log Mean Driving Force Method Solution Gm=1.0 kg/m2.s
Lm=1.6 kg/m2.s x2 = 0
y2 = 0.00015 2
Given:
Reduced to 1% 1 y1 = 0.015 Gm=1.0 kg/m2.s
x1 = ? Lm=1.6 kg/m2.s
Cross sectional A of column = 1 m2 Mwt air = 29 Mwt acetone = 58 Mwt H2O = 18 HOG = 0.3 m y* = 1.75 x
Log Mean Driving Force Method Solution The given mass flow rate in kg/m2.s. Need to convert into kmol /s. Gm
Gm in mass unit Area M wt
kg 2 m s 1m 2 0.03448 kmol Gm kg s 29 kmol 1.0
Find slope of the graph. L L x x1 y1 ope ratingline G G L 0.08889 kmol / s slope 2.578 G 0.03448 kmol / s y
Lm
Lm in mass unit Area M wt
kg 2 m s 1m 2 0.08889 kmol Lm kg s 18 kmol 1.6
Log Mean Driving Force Method Solution Calculate the mol ratio for each location. Location 1
Location 2
y1
A 0.015 0.0152 Gm 1 0.015
y2
A 0.00015 0.00015 Gm 1 0.00015
x1
A ? Lm
x2
A 0 0 Lm 1 0
Need to find x2 in order to plot operating line. L L x x1 y1 G G y 2.578 x C y
Log Mean Driving Force Method Solution Use coordinate at location 2 in order to find C value. y 2.578 x C 0.00015 2.5780 C C 0.00015
Operating line, y 2.578x 0.00015
Insert value at location 1 in order to find x1 value y1 2.578 x1 0.00015 0.015 2.578 x1 0.00015 x1
0.015 0.00015 0.00576 2.578
Log Mean Driving Force Method Solution Need to find y1* and y2* using equilibrium line given in question.
y1* Hx1
y * 1.75 x
y1* 1.75 x1 1.750.00576 y1* 0.010 y2* 1.75 x2 1.750 y2* 0
Log Mean Driving Force Method Solution Calculate NOG. N OG
y1 y2
ln y y *
y y y y 0.0152 0.010 0.00015 0 ln y y y y 0.0152 0.010 ln ln 0.00015 0 y y 5.05 10 ln y y 1.4242 10 *
*
*
1
2
*
1
*
2
3
*
5.2 10 3 ln 0 . 00015
3
Log Mean Driving Force Method Solution Calculate NOG. N OG
y1 y2 0.0152 0.00015 10.57 11 stages
ln y y *
1.4242 103
Height of absorber, Z
Z H OG N OG
Z 0.3m 11 Z 3.3m
Colburns Method Objective to evaluate: y1
y2
dy y y*
Assume mol fraction are so small fraction; mol fraction = mol ratio
Gm y1 y2 Lm x1 x2 Also, y*=Hx, but instead of H, use m. So y*=mx
y*
mGm y y2 Lm
If we substitute, we get y1
y2
dy mGm y y2 y Lm
Solution as log function and represented as a nomogram (graphical solution)
Colburns Method Example An acetone – air mixture containing 0.015 mol fraction of acetone will be reduced to 1% of this value by countercurrent absorption in fresh pure water in packed tower. The gas flow rate is 1.0 kg/m2.s and the liquid flow rate is 1.6 kg/m2.s. For the system, Henry’s law holds and y* = 1.75 x where y* is the mol fraction of acetone in the vapor in equilibrium with x mol fraction in liquid. Calculate the height of the tower / absorber if HOG = 0.3 m.
Data given: Cross sectional A of column = 1 m2 Mwt air = 29 Mwt acetone = 58 Mwt H2O = 18
Colburns Method Solution Gm=1.0 kg/m2.s
Lm=1.6 kg/m2.s x2 = 0
y2 = 0.00015 2
Given:
Reduced to 1% 1 y1 = 0.015 Gm=1.0 kg/m2.s
x1 = ? Lm=1.6 kg/m2.s
Cross sectional A of column = 1 m2 Mwt air = 29 Mwt acetone = 58 Mwt H2O = 18 HOG = 0.3 m y* = 1.75 x
Colburns Method Solution The given mass flow rate in kg/m2.s. Need to convert into kmol /s. Gm
Gm in mass unit Area M wt
kg 2 m s 1m 2 0.03448 kmol Gm kg s 29 kmol 1.0
Find mGm/Lm and y1/y2 m
Gm 0.03448 1.75 0.6788 Lm 0.08889
y1 0.015 100 y2 0.00015
Lm
Lm in mass unit Area M wt
kg 2 m s 1m 2 0.08889 kmol Lm kg s 18 kmol 1.6
Colburns Method
Colburns Method Solution From nomogram, NOG = 10.6 = 11 Height of absorber, Z
Z H OG N OG
Z 0.3m 11 Z 3.3m
Plate – tower → Description and Design
INTRODUCTION TO PLATE – TOWER
DETERMINATION STAGES OF TOWER
MINIMUM LIQUID FLOW RATE TO OBTAIN SPECIFIC SEPARATION
Introduction to Plate – tower Bubble – cap columns or sieve trays, are sometimes used for gas absorption.
Application of plate tower – particularly when the load is more than can be handled in a packed tower about 1 m diameter; and - when there is any likelihood of deposition of solids which would quickly choke a packing. Plate tower are particularly useful when the liquid rate is sufficient to flood a packed tower.
Determination stages of tower It will be assumed that dilute solutions are used so that mole fraction and mole ratio approximately equal. A material balance for the absorbed component from the bottom to a plane above plate n will give:
Gm yn Lm xs Gm ys 1 Lm xn 1 yn
Lm L xn 1 ys 1 m xs Gm Gm
Operatingline
Operating line also describes such a line passes through two point, 1 (top of tower) and 2 (bottom of tower)
Determination stages of tower Example 1 A bubble – cap absorption column is to be used to absorb ammonia, NH3 by using water. A gaseous mixture containing 20.5 mol% NH3 and 79.5 mol% air enters the bottom of the absorption tower. 60.5 kmol of gaseous NH3 enters the tower per hour while 5500 kg aqueous NH3 solution containing 0.1% by mass NH3 enters the top of the tower per hour. The column operates at the atmospheric pressure and at a constant temperature of 30OC. It is desired to absorb 95% of the entering gas NH3. Assume that the effect of water vapor in the gases is negligible.
Determination stages of tower Example Determine: The molar flow rate of entering gaseous mixture
The molar flow rate of the raffinate The molar flow rate of the extract The mol ratio of NH3 in the raffinate and extract The number of ideal stages required The equilibrium data given as follows: Mol NH3 mol water
0.000
0.053
0.111
0.177
0.250
Mol NH3 mol air
0.000
0.044
0.089
0.159
0.280
Determination stages of tower Solution Lm
Gm
Feed of NH3 aq in= 5500 kg/h
y2 = 0.00015
xA2 = 0.001 kg A/kg
2
xB2 = 0.999 kg B/kg
95% removal T=30OC P=1 atm 1 x1 = ? Gm yA1 = 0.205 kmol A/kmol yB1 = 0.795 kmol B/kmol Feed of NH3 in= 60.5 kmol/h
Lm
A = NH3 B = Air C = H2O
Determination stages of tower Solution Determine:
The molar flow rate of entering gaseous mixture, G1 From Gaseous mixture inlet (at point 1),Feed NH3 in = 60.5 kmol/h
kmol A kmol A G1 60.5 kmol h kmol A 1 kmol G1 60.5 h 0.205 kmol A kmol A G1 295.12 h 0.205
Determination stages of tower Solution Determine: The molar flow rate of the raffinate, G2 From information, 95% of the entering NH3 is being absorbed. Unabsorbed NH3 = 0.05 60.5
kmol kmol 3.025 h h
The molar flow rate of the raffinate, G2 = Gm + unabsorbed NH3 Gm 0.795
kmol B kmol B kmol kmol B G1 0.795 295.12 234.62 kmol kmol h h
G2 Gm unabsorbe dNH 3 234.62 3.025 G2 237.645
kmol h
kmol h
Determination stages of tower Solution Determine: The molar flow rate of the extract, L1 Absorbed NH3 = 0.95 60.5
kmol kmol 57.475 h h
From information at L2, feed NH3 aq. in = 5500 kg/h Lm 0.999
kg C kg C kg kg C L2 0.999 5500 5494.5 kg kg h h
LA2 0.001
kg A kg A kg kg A L2 0.001 5500 5.5 kg kg h h
Convert Lm and LA2 in kmol/h unit. Given: Mwt NH3 = 17
Mwt H2O = 18
Determination stages of tower Solution Lm 5494.5 LA2 5.5
kg C 1 kmol C kmol C 305.25 h 18 kg C h
kg A 1 kmol A kmol A 0.324 h 17 kg A h
The molar flow rate of the extract, L1 L1 Lm LA2 Absorbe dNH 3 L1 305.25 0.324 57.475 L1 363.049
kmol h
kmol h
Determination stages of tower Solution Determine:
The mol ratio of NH3 in all stream
Location 1 Gas Feed stream, G1 Mol of NH3 = 60.5 kmol/h Mol of Air, Gm = 234.62 kmol/h
Extract stream, L1 Mol of NH3 = 0.324 + 57.475 kmol/h = 57.799 kmol/h Mol of Water, Lm = 305.25 kmol/h
mol of NH 3 mol of Air 60.5 kmol Y1' 234.62 kmol Y1' 0.258 Y1'
Coordinate for Location 1 = (X1’, Y1’) = (0.189,0.258)
mol of NH 3 mol of wate r 57.799 kmol X 1' 305.25 kmol X 1' 0.189 X 1'
Determination stages of tower Solution Location 2 Raffinate stream, G2
Liquid feed stream, L2
Mol of NH3 = 3.025 kmol/h
Mol of NH3 = 0.324 kmol/h
Mol of Air, Gm = 234.62 kmol/h
Mol of Water, Lm = 305.25 kmol/h
mol of NH 3 Y mol of Air 3.025 kmol Y2' 234.62 kmol Y2' 0.013 ' 2
Coordinate for Location 2 = (X2’, Y2’) = (0.001,0.013)
mol of NH 3 mol of wate r 0.324 kmol X 2' 305.25 kmol X 2' 0.001 X 2'
Determination stages of tower Solution Plot the equilibrium line and the operating line in order to calculate number of stages No of theoretical stages = 5 theoretical stages
Determination stages of tower 0.3 0.275 0.25 0.225 0.2
y
0.175 0.15 0.125 0.1 0.075 0.05 0.025 0 0
0.025 0.05 0.075
0.1
0.125 0.15 0.175 x
0.2
0.225 0.25 0.275
0.3
Determination stages of tower Example 2 A bubble – cap absorption column is to be used to absorb ammonia, NH3 by using water. A gaseous mixture containing 15 mol% NH3 and 85 mol% air enters the bottom of the absorption tower. 57 mol of gaseous NH3 enters the tower per hour, while 3.7 kg of pure water enters the top of the tower per hour. The column operates at the atmospheric pressure and at a constant temperature of 30OC. It is desired to absorb NH3 such as only 2.5% NH3 leaves the tower with air. Determine the number of theoretical plates required for the above process. Equilibrium data is given as in Example 1. Assume that the effect of water vapor in the gases is negligible.
Minimum Liquid Flow rate Min. flow rate = Infinite number of stages = Minimum reflux in distillation.
y
as L decreases, line approaching equilibrium line
equilibrium line
x Minimum value of flow rate is where: Exit liquid [ ] = Equilibrium [ ] or… Inlet gas [ ] = Equilibrium [ ]
