liquid gas absorption process
TRANSCRIPT
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1
Gas-Liquid Separation
Absorption Process
Part 01
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+ 2Content
Types of Separation Processes and Methods Equilibrium Relation Between Phases
Single and Multiple Equilibrium Contact Stages
Mass Transfer Between Phases
Continuous Humidification Processes
Absorption on plate and Packed Tower
Absorption and Concentrated Mixtures in Packed
Tower
Estimation of Mass Transfer Coefficient in Packed Tower
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+ 3Types of Separation Processes and Methods
In order to separate or remove one or more of the component
from its original mixture, it must be contact with anotherphase to allow solute to diffuse from one phase to the others
phase.
During the contact of the two phase, the component of the
original mixture redistribute themselves between the two
phases. The phases are then separated by simple physical
methods.
By choosing the proper conditions and phases, one phase is
enriched while the other phase is depleted in one or more
components.
The two phase pair can be:
1. Gas (vapor)-liquid :absorption, distillation, etc
2. Gas –solid :adsorption, etc
3. Liquid-liquid :liquid-liquid extraction, etc
4. Liquid- solid :leaching, crystallization, etc
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+ 4Gas-Liquid Separation
Absorption
A solute A or several solutes from a gas phase are absorbed into aliquid phase.
involves molecular and turbulent diffussion or mass transfer of solute A through a stagnant, nondifussing gas B into a stagnantliquid C
e.g: absorbing NH3 (A) from air (B) using liquid water (C).
Stripping/desorption
reverse of absorption: separating a solute A or several solutesfrom a liquid phase by contacting the liquid with a gas phase.
E.g. removal of volatile component of the oils by passing out withthe steam.
Humidification
When the gas is pure air and the liquid is pure water
transfer of water vapor from liquid water into pure air. Dehumidification
removal of water vapor from air.
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+ 5Equilibrium Relation Between PhasesGas-Liquid Equilibrium – Henry’s Law
Henry’s Laws:
where,
p A = partial pressure of component A (atm).
H = Henry’s law constant (atm/mol
fraction).
H’ = Henry’s law constant (mol frac.gas/mol
frac.liquid). = H/P.x A = mole fraction of component A in
liquid.
y A = mole fraction of component A in gas =
p A /P.
P = total pressure (atm).
H’ depend on total pressure, whereas H does
not.
At low concentration – a straight line on plot
p A vs x A .
A A Hx p A A x H y '
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+ 6Single Stage Equilibrium Contact
Single-stage process
two different phases (liquid & gas for absorption) are brought into contact
& then separated.
various component diffuse and redistribute themselves between two
phase
long enough, equilibrium is reached.
Two exit stream L1 andV1 leave in equilibrium with each other.
Gas phase contains solute A & inert gas B
Liquid phase contains solute A & inert liquid (solvent) C.
Inert gas is insoluble in the solvent and the solvent does not vaporize to
the gas phase.
L1 ,xA1Lo ,xAo
V2, yA2V1 , yA1
SingleStageLiquid phase
inletLiquid phase
outlet
Gas phaseinlet
Gas phaseoutlet
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+ Single Stage Equilibrium Contact for Absorption
Total material balance
Component A balance on single stage equilibrium contact for
absorption
Both L’ and V’ are constant and usually is known
To solve the value x A1 and y A1, we use a Henry’s Law relationship
Remember the unit H’ is mol frac.gas/mol frac.liquid
7
Lo x Ao +V 2 y A2 = L1 x A1 +V 1 y A1 = Mx AM
1
1
1
1
2
2
1'1'1'1' A
A
A
A
A
A
Ao
Ao
y
y
V x
x
L y
y
V x
x
L
A A x H y '
Lo +V 2 = L1 +V 1 = M
L1 ,xA1Lo ,xAo
V2, yA2V1 , yA1
SingleStageLiquid phase
inletLiquid phase
outlet
Gas phaseinlet
Gas phaseoutlet
V = gas flow rate
V’= inert/pure gas (B) flow rate = V(1-y A )
L = liquid flow rate
L’= inert/pure liquid (C) flow rate = L(1-x A )
x= mole fraction in liquid phase
y = mole fraction in gas phase
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+ 8
A gas mixture at 1.0 atm pressure abs containing air and CO2is contacted in a single-stage mixer continuously with pure
water at 293 K. The two exit gas and liquid streams reach
equilibrium. The inlet gas flow rate is 100 kg mol/h, with a
mole fraction of CO2 of y A2 = 0.20. The liquid flow rate
entering is 300 kg mol water/h. calculate the amounts and
compositions of the two outlet phases. Assume that waterdoes not vaporize to the gas phase.
Example 10.3-1
L1= ?xA1 =?
L0= 300 kg/h
xAo =0
V2= 100 kg/hyA2 =0.20
Single-stageV1= ?
yA1 =?
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+ 9Solution Example 10.3-1
The inert water flow is L’ = L0 = 300 kg mol/h.
The inert air flow V’ is obtained from, V’ = V(1-y A ). Hence, the inert air flowis
V’ = V2 (1-y A2 )
= 100 (1-0.20)
= 80 kg mol/h .
Substituting into equation to make a balance on CO2 (A).
……….(1)
1
1
1
1
180
1300
20.01
20.080
01
0300
A
A
A
A
y
y
x
x
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+ 10
At 293 K, the Henry’s law constant from Appendix A.3 is
H = 0.142 x 104 atm/mol frac.
Then
H’ = H/P = 0.142 x 104 /1.0 = 0.142 x 104 mol frac. gas/mol frac.liquid.
Substituting into,
y A1 = 0.142 x 104 x A1 ………(2)
Solve simultaneous equation.
x A1 = 1.41 x 104 and y A1 = 0.20
To calculate the total flow rates leaving,
hkgmol x
L
L A
/ 3001041.11
300
1 41
'
1
hkgmol y
V V
A
/ 10020.01
80
1 1
'
1
Solution Example 10.3-1
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+ 11
Countercurrent multiple-contact stages.
more concentrated product.
total number of ideal stages = N.
Component B&C may or may be not be somewhat miscible in each other.
two streams leaving a stage in equilibrium with each other. Total overall material balance: L 0 + V N+1 = L N + V 1 = M
Component A balance: L 0 x 0 + V N+1 y N+1 = L N x N + V 1 y 1 = Mx M
For the first n stages: L 0 x 0 + V n+1 y n+1 = L n x n + V 1 y 1
Operating line:
An operating line is an important material-balance equation because it relatesthe concentration yn+1 in the V stream with xn in the L stream passing it.
Slope= Ln/Vn+1 in operating line is varies if the L and V streams vary from stage tostage
1 2 n N
L0
L1
L2
Ln - 1 L n L N - 1 L N
V N + 1
V N
V n + 1
V n
V 3
V 2
V 1
Countercurrent Multiple Contact Stage
yn+1 = LnV n+1
xn +V
1 y
1 − L0 x0V n+1
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+ 12Countercurrent Multiple Contact Stage – Number of Ideal Stages
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Graphical calculation for determining number of ideal stage,
N:
1) Plot y A vs x A .
2) Draw operating line.
3) Draw equilibrium line (Henry’s law).
4) Stepping upward (or downward) until yN+1 (or y1) is reached5) N = number of steps/trays
Dilute system (
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+ 14Example 10.3-2
It is desired to absorb 90% of the acetone in a gas containing
1.0 mol % acetone in air in a countercurrent stage tower. The
total inlet gas flow to the tower is 30.0 kg mol/h and the total
inlet pure water flow to be used to absorb the acetone is 90
kg mol H2O/h. The process is to be operate isothermally at
300 K and a total pressure of 101.3 kPa. The equilibrium
relation for the acetone (A) in the gas-liquid is y A = 2.53x A .Determine the number of theoretical stages required for this
separation.
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+ 15Solution Example 10.3-2
Given values are y AN+1 = 0.01, x A0 = 0,VN+1 = 30.0 kg mol/h, and L0 =L’= 90.0 kg mol/h.
Amount of entering acetone = y AN+1 VN+1= 0.01(30.0)
= 0.3 kg mol/h.
Entering air,V’ = (1- y AN+1 )VN+1= (1-0.01)(30.0)
= 29.7 kg mol air/h
Acetone leaving in V1 = 0.10(0.30) = 0.03 kg mol/h.
Acetone leaving in L1
= 0.90(0.30) = 0.27 kg mol/h. (10% avetone enter is absorbed)
V1 = 29.7 + 0.03 = 29.73 kg mol air + acetone/h.
y1 = (0.030/29.73) = 0.00101
LN = 90.0 + 0.27 = 90.27 kg mol water + acetone /h.
xN = (0.27/90.27) = 0.0030.
Since the flow of liquid varies only slightly from L0 = 90.0 at the inlet to LN = 90.27 at the the outlet and V from 30.0 to 29.73, the slope Ln /Vn+1 of the operating line is essentially constant. This line is plotted,and the equilibrium relation y A = 2.53x A is also plotted.
Starting at point y A1 ,x A0 the stages are drawn. About 5.2 theoretical stages are required.
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+ 16Solution Example 10.3-2
0 0 0.001 0.002 0.003 0.004
0.004
0.008
0.012
xANxA0
yA1
yAN+1
Equilibrium line
Operating line
3
2
1
4
5
Mole fraction acetone in water, xA
Mole fraction acetone in air, yA
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+ 17 Analytical Equations For Countercurrent StageContact : Kremser equations
Kremser equations
calculate the number of ideal stages.
valid only when operating & equilibrium lines are straight.
v, L have a constant value
Absorption Stripping
When A = 1
N = y N +1 − y1 y
1 −mx
0
N = In
x0 − ( y N +1 / m) x N − ( y N +1 / m) 1− A( ) + A
In(1 / A) N =
In y N +1 −mx0 y1 −mx01− 1 A +
1 A
InA
N = x0 − x N x N − y N +1 / m
Procedure (for varying A):
1.Calculate A 1 = L0/mV1 at L0 & V1.
2.Calculate A N = LN/mVN+1 at LN &VN+1.3.Calculate geometric average area, A ave. =
4. Calculate N.
m = slope of equilibrium line.
A = absorption factor = constant = L/(mV).
L, V = molar flow rates
N A A
1
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+ 18Example 10.3-3
Repeat Example 10.3-2 but use the Kremser analytical equation
for countercurrent stage processes.
Solution
At one end of the process at stages 1,V 1 = 29.73 kg mol/h y A1 = 0.001001
L 0 = 90.0, and x A0 = 0.
Also, the equilibrium relation is y A = 2.53x A where m = 2.53.
Then,
20.173.2953.2
0.90
1
0
1 mV
L
mV
L
A
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+ 19Solution Example 10.3-3
At stage N,
VN+1 =30.0, y AN+1 = 0.01,LN = 90.27, and x AN = 0.0030.
The geometric average,
Then,
This compares closely with 5.2 stages using graphical method
19.10.3053.2
27.90
1
N
N N
mV
L A
195.119.120.01 N A A A
04.5)195.1log(
195.1
1
195.1
11
)0(53.200101.0
)0(53.201.0log
N
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+Next Part
Types of Separation Processes and Methods
Equilibrium Relation Between Phases
Single and Multiple Equilibrium Contact Stages
Mass Transfer Between Phases
Continuous Humidification Processes
Absorption on plate and Packed Tower
Absorption and Concentrated Mixtures in Packed Tower
Estimation of Mass Transfer Coefficient in Packed Tower
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21
Gas-Liquid Separation
Part 02
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+Next Part
Types of Separation Processes and Methods Equilibrium Relation Between Phases
Single and Multiple Equilibrium Contact Stages
Mass Transfer Between Phases
Continuous Humidification Processes Absorption on plate and Packed Tower
Absorption and Concentrated Mixtures in Packed
Tower
Estimation of Mass Transfer Coefficient in PackedTower
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+Subtopics:
Mass Transfer Between Phases
Introduction
Film Mass Transfer Coefficient and Interface Concentration :
Equimolar Counterdiffusion
Film Mass Transfer Coefficient and Interface Concentration :
Diffusion of A through Stagnant & Nondiffusing B
Overall Mass-Transfer Coefficients:
Equimolar Counterdiffusion/Diffusion in Dilute Solutions
Overall Mass-Transfer Coefficients:
Diffusion of A Through Stagnant & Nondiffusing B
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+Mass Transfer Between Phases
y AG - average or bulk mole fraction of A in gas phase
x AL - average or bulk mole fraction of A in liquid phase
y Ai & x Ai - equilibrium mole fraction at interface
25
Distance from interface
NA
yAGyAi
interface
gas-phase mixture
of A in gas G
liquid-phase solution
of A in liquid L.
xAixAL
Concentration profile of solute A diffusing through two phases.
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+Film Mass Transfer Coefficient and InterfaceConcentration : Equimolar Counterdiffusion
k’y -gas phase mass transfer coefficient (kgmol/s.m2.mol frac.)
k’x -liquid phase mass transfer coefficient (kgmol/s.m2.mol frac.)
Point P is bulk phase compositions y AG and x AL
Point M is interface concentration x Ai and y Ai
26
)(')(' AL Ai x Ai AG y A x xk y yk N
)(
)(
'
'
Ai AL
Ai AG
y
x
x x
y y
k
k
If two film coefficient k’x and k’yare known, the interface
composition can be
determined by drawing line
PM with slope –k’x/k’yintersecting the equilibrium
line
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+ Film Mass Transfer Coefficient and InterfaceConcentration : Diffusion of A through Stagnant &
Nondiffusing B
Trial and error need to get the slope. For first trial assume (1-y A )iMand (1-x A )iM equal to 1.
27
)]1 /()1[(
)1()1()1(
)]1 /()1[(
)1()1()1(
)1(
'
)1(
')()(
Ai AL
Ai AL
iM A
AG Ai
AG Ai
iM A
iM A
x
x
iM A
y
y
AL Ai x Ai AG y A
x x In
x x x
y y In
y y y
x
k k
y
k k
x xk y yk N
)(
)(
)1 /('
)1 /('
)()1(
')()1(
'
Ai AL
Ai AG
iM A y
iM A x
AL Ai
iM A
x Ai AG
iM A
y A
x x
y y
yk
xk
x x xk y y
yk N
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+ Example 10.4-1The solute A is being absorbed from a gas mixture of A and B in a wetted-wall
tower with the liquid flowing as a film downward along the wall. At certain point
in the tower the bulk gas concentration y AG = 0.380 mol fraction and the bulk
liquid concentration is x AL = 0.1. The tower is operating at 298 K and 1.013 x 105
Pa and the equilibrium data are as follows:
x A y A x A y A
0 0 0.20 0.131
0.05 0.022 0.25 0.187
0.10 0.052 0.30 0.265
0.15 0.087 0.35 0.385
The solute A diffuse through stagnant B in the gas phase and then through a
nondiffusing liquid. Using correlations for dilute solutions in wetted-wall
towers, the film mass-transfer coefficient for A in the gas phase is predicted as
ky = 1.465 x 10-3 kg mol A/s.m2 mol frac. And for the liquid phase as kx = 1.967
x 10-3 kg mol A/s.m2 mol frac. Calculate the interface concentrations y Ai and x Ai
and the flux N A .
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+Overall Mass-Transfer Coefficients andDriving Force
Film or single phase mass transfer coefficient k’y and k’xor ky and kx are often difficult to measure
As a result, overall mass transfer coefficient K’y and Kx’ are
measured based on gas phase or liquid phase.
K’y - overall gas phase driving force (kgmol/s.m2.mol frac.)
K’x - overall liquid phase driving force (kgmol/s.m2.mol frac.)
y* A - mole fraction that equilibrium with x AL
X* A - mole fraction that equilibrium with y AG
29
N A = K ' y ( y AG − y A* ) = K ' x ( x A* − x AL )
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+ Overall Mass-Transfer Coefficients:Equimolar Counterdiffusion/Diffusion in Dilute Solutions
30
Gas phase controlling (i.e major resistance in gas
phase)- m’ is small, so that the equilibrium curve is
almost horizontal, a small value of yA in the gas willgive a large value of xA in equilibrium in liquid. A is
very soluble in liquid phase
Liquid phase controlling (i.e major resistance in
liquid phase) - m” is large and A is very insoluble
in liquid phase
N A = K ' y ( y AG − y A*
) = K ' x ( x A* − x AL )
1
K ' y = 1
k y ' + m '
k x 'm ' =
y Ai − y A*
x Ai − x AL1
K ' x= 1m"k y '
+ 1k x '
m" = y AG − y Ai x A
* − x Ai
'
1
'
1
y y k K
'
1
'
1
x x k K
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+ Overall Mass-Transfer Coefficients:Diffusion of A Through Stagnant & Nondiffusing B
31
)()1(
)()1(
''
AL A
M A
x A AG
M A
y
A x x x
K y y
y
K N
)]1 /()1[()1()1()1(
)]1 /()1[()1()1()1(
)1(
'
)1(
'
*
*
**
*
*
**
A AL
A AL M A
AG A
AG A M A
M A
x
x
M A
y
y
x x In x x x
y y In y y y
x
K K
y
K K
iM A xiM A y M A y xk
m
yk yK )1 /('
'
)1 /('
1
)1 /('
1
*
iM A xiM A y M A x xk yk m xK )1 /('
1
)1 /('"
1
)1 /('
1
*
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+Example 10.4-2
Using the same data as in Example 10.4-1, calculate the overall
mass transfer coefficient K’y and the percent resistance in the
gas and liquid films. Do this for the case of A diffusing through
stagnant B.
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+Subtopics:
Continuous Humidification Processes
Water-Gas Interface for Water Cooling Tower
Common Term in Humidification
Operating Line for Water Cooling Tower
Tower Height For Water-Cooling
Design of Water Cooling Tower Using Film Mass Transfer
Coefficient
Design of Water Cooling Tower Using Overall Mass Transfer
Coefficient
Minimum Value of Air Flow
Design of Water Cooling using Height of Transfer Unit
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+Continuous Humidification Process
When a relatively warm liquid is brought into direct contact with a gas that is
unsaturated, some of the liquid is vaporized to the gas phase.
This done for the following purpose
Humidifying of air for control the moisture content of air in drying or air conditioning
Dehumidifying air where cold water condenses some water vapor from air.
Water cooling where water is evaporated to air to cool the warm water.
Typical water cooling tower
Warm water flows counter currently to an air stream. The warm water enters the top of a packed tower and cascades down through the
packing,leaving at the bottom.
Air enters at the bottom of the tower and flows upward through the descending water
by the natural draft or by the action of a fan.
The water is distributed by troughs and overflows to cascade over slat gratings or
packing that provide large interfacial areas of contact between the water air air in
the form of droplets and film of water. The tower packing often consists of slats of wood or plastic or of a packed bed
In humidification and dehumidification, the gas phase resistance controls the
rate of the mass transfer.
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+ Water-Gas Interface for Water CoolingTower
Sensible heat flow from liquid to the interface= sensible heat flow in
the gas phase + latent heat (vaporization) flow in the gas.
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+ Common Term in HumidificationChapter 9.3
Humidity (H) of an air-water vapor is the kg of water vapor
contained in 1 kg of dry air.
Saturated humidity (Hs)- air in which the water vapor is equilibrium
with liquid water at given P & T. Partial pressure of water vapor inair-water mixture is equal to the vapor pressure p As of pure water at
given T
Percentage humidity (Hp)
Percentage relative humidity (HR)
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+Common Term in Humidification
Dew point
the temperature at which a given mixture of air and water vapor would be saturated.
or temperature at which vapor begins to condense when the gas
phase is cooled at constant pressure.
Humid heat cS amount of heat required to raise the temperature of 1 kg of dry air
plus the water vapor present by 1 K.
cS (kJ/kg dry air.K) = 1.005 + 1.88H (SI)
where, cP water(v) = 1.88 kJ/kg water vapor. K and cP air = 1.005
kJ/kg dry air. K
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+Common Term in Humidification
Humid volume, vH - total volume (m3) of 1 kg of dry air plus
the vapor it contains at 1 atm abs pressure and the given gastemp.
vH (m3/kg dry air) = (2.83 x 10-3 + 4.56 x 10-3 H) T (K).
Total enthalpy of an air-water mixture, H Y - the total enthalpyof 1 kg of air plus its water vapor. Sensible heat of the air-
water vapor mixture plus the latent heat.
H Y (kJ/kg dry air) = (1.005 + 1.88 H) (T ºC-0) + 2501.4
where,Tref for both components = 0 ºC
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+ Humidity Chart39
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+ Operating Line for Water CoolingTower
Assumption:
1. Process carried out adiabatic
2. L is constant
3. cL is constant at 4.187 x 103 J/kg. K
Slope= LcL/G
G = dry air flow, kg/s.m2.
L = water flow, kg water/s.
cL = heat capacity of water, assumed constant at 4187 kJ/kg.K.
TL= temperature of water, ºC or K.
Hy= enthalpy of air-water vapor mixture, J/kg air.
= cs (T-T0) + H0 = (1.005 + 1.88H)103 (T-0) + 2.501x106H
H = humidity of air, kg water/kg dry air – Humidity chart in Fig. 9.3-2pg. 568.
40
)()( 11 L L L y y T T Lc H H G
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+ 41Tower Height For Water-Cooling
Tower height for water-cooling
z = column height
MB = molecular weight of air
a = m2 of interfacial area per m3 volume of packed section
kG = gas phase mass transfer coefficient ( kgmol/s. m2)
kGa = volumetric coefficient in kgmol/s.m3 volume
P = pressure in atm
Slope of line from interface composition to point in equilibrium line
2
1
Hy
Hy y yi
y
G B H H
dH
aPk M
G z
Li
y yi
BG
L
T T
H H
PaM k
ah
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+ Temperature Enthalpy Diagram AndOperating Line For Water- Cooling
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+ Enthalpies of Saturated Air-Water VaporMixture (0C Base Temperature)
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+ Design of Water Cooling Tower UsingFilm Mass Transfer Coefficient
1. Draw equilibrium line ( Data from table 10.5-1) in H-versus-T plot
2. Draw operating line with slope LcL/G.
3. Determine Hyi-Hy by plot a line from P (TL, Hy) to M (Ti, Hyi). The slope of thisline –hLa/kGaMBP. Do for various P point between (TL1, Hy1) to (TL2, Hy2).
4. Use numerical or graphical integration to obtain the value of integration in
height tower equation. For example using graphical method, plot 1/(Hyi-Hy)
versus Hy.
44
2
1
Hy
Hy y yi
y
G B H H
dH
aPk M
G z
Li
y yi
BG
L
T T
H H
PaM k
ah
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+ Design of Water Cooling Tower UsingOverall Mass Transfer Coefficient
Often overall mass transfer coefficient KGa (kgmol/s.m3.Pa) is available. Then
to determine tower height, z
1. Draw equilibrium line ( Data from table 10.5-1) in H-versus-T plot
2. Draw operating line with slope LcL/G.
3. Determine H*y-Hy by plot a line from P (TL, Hy) to R (TL, H*y). The slope of thisline –hLa/kGaMBP. Do for various P point between (TL1, Hy1) to (TL2, Hy2).
4. Use numerical or graphical integration to obtain the value of integration inheight tower equation. For example using graphical method, plot 1/(Hyi-Hy)
versus Hy.
45
Li
y yi
BG
L
T T
H H
PaM k
ah
2
1 *
Hy
Hy y y
y
G B H H
dH
aPK M
G z
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+Example 10.5-1
A packed countercurrent water-cooling tower using a gas flow
rate of G = 1.356 kg dry air/s. m2 and a water flow rate of L =1.356 kg water/s. m2 to cool the water from TL2 = 43.3 ºC to TL1= 29.4 ºC. The entering air at 29.4 ºC has a wet bulb
temperature of 23.9 ºC . The mass-transfer coefficient kG a is
estimated as 1.207 x 10-7 kg mol/s.m3.Pa and hL a / kG aMBP as
4.187 x 104
J/kg.K . Calculate the height of packed tower z. Thetower operates at a pressure of 1.013 x 105 Pa.
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+ 47Solution Example 10.5-1
Following the step outlined, the enthalpies from the saturated air-water vapor
mixtures from Table 10.5-1 are plotted in Fig. 10.5-4.
The inlet air at TG1 = 29.4 ºC has a wet bulb temperature of 23.9 ºC . The
humidity from the humidity chart is H1 = 0.0165 kg H2O/kg dry air.
Substituting into Eq.(9.3-8),
Hy1 = cs (T-T0) + H0 = (1.005 + 1.88H) (T-T0) + H0
Hy1 = (1.005 + 1.88 x 0.0165)103 (29.4-0) + 2.501 x 106 (0.0165).
= 71.7 x 103 J/kg.
The point Hy1 = 71.7 x 103 and TL1 = 29.4 ºC is plotted.
Then substituting into Eq. (10.5-2) and solving,
G(Hy2 –Hy1) = LcL (TL2 – TL1)
1.356 (Hy2 – 71.7 x 103) = 1.356 (4.187 x 103) (43.3 – 29.4).
Hy2 = 129.9 x 103 J/kg dry air.
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+ 48Solution Example 10.5-1
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+ 49
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+ 50
+ Mi i V l f Ai Fl 51
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+ Minimum Value of Air Flow The air flow G is not fixed but must be set for the design of the cooling
tower.
For a minimum value of G, the operating line MN is drawn through the
point Hy1 and TL1 with a slope that touches the equilibrium line at TL2,point N.
If the equilibrium line is quite curved, line MN could become tangent tothe equilibrium line at a point farther down the equilibrium line thanpoint N.
For the actual tower, a value of G greater than Gmin must be used. Often,a value of G equal to 1.3 to 1.5 times Gmin is used.
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+ D i f W t C li i H i ht f 52
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+ Design of Water Cooling using Height of Transfer Unit
Using Film Mass Transfer Coefficient
Using Overall Mass Transfer Coefficient
52
2
1
2
1
Hy
Hy y yi
y
G
G B
G
Hy
Hy y yi
y
G B
H H
dH N
aPk M
G H
H H
dH
aPk M
G z
2
1 *
2
1 *
Hy
Hy y y
y
OG
G B
OG
Hy
Hy
y y
y
G B
H H
dH N
aPK M
G H
H H
dH
aPK M
G z
z= height of transfer unit
X number of transfer unit
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+
53
Gas-Liquid SeparationPart 03
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+Next Part
Types of Separation Processes and Methods
Equilibrium Relation Between Phases
Single and Multiple Equilibrium Contact Stages
Mass Transfer Between Phases
Continuous Humidification Processes
Absorption in Plate and Packed Tower Pressure Drop & Flooding in Packed Towers
Absorption of Dilute Gas Mixtures in Packed Tower
Absorption of Concentrated Gas Mixtures in
Packed Tower Design of Packed Towers Using Transfer Units
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+Equipment for Absorption & Distillation
1. Plate (tray) tower
Sieve tray
Valve tray
Bubble cap tray
2. Structured packing
3. Packed tower
Raschig ring
Lessing ring
Berl saddle
Pall ring
etc
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+Tray/Plate Tower
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+Tray (Plate) Tower
Sieve tray
Valve tray
57
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+Tray (Plate) Tower
Bubble cap tray
58
+ S d P ki59
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+ Structured Packing59
+ S d P ki60
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+ Structured Packing
Corrugated structured packing
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+Packed Tower
+ Packed Tower 62
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+ Packed Tower
+ P D & Fl di i P k d T 63
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+ Pressure Drop & Flooding in Packed Towers
Flooding velocity – upper limit of gas flow rate
At low gas velocities the liquid flow downward through the
the packing, essentially uninfluenced by the upward gas flow
As the gas flow rate 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 start to
hinder the liquid downflow, and local accumulations or pools
of liquid start to appear in the packing. The pressure drop of
the gas starts to rise at a faster rate
As the gas flow arte is increased, the liquid holdup or
accumulation increase. At the flooding point, the liquid no
longer flow down through the packing and is blown out withthe gas.
The optimum economic gas velocity is about 0ne-half or
more of the flooding velocity
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+ Pressure Drop in Random Packing
Note: this capacity parameters is not dimensionless and that onlythese unit (english) should be used
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+ Pressure Drop in Structured Packing
+ Packing Factors (F ) for Random and 66
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+ Packing Factors (Fp) for Random andStructured packing
+ Flooding Pressure Drops in Packed and 67
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+ Flooding Pressure Drops in Packed andStructured Packing
Can be used for packing factors from 9 up to 60. It predicts all the
data for flooding within +- 15% and most for +-10% for packing factor of 60 or higher, this equation is not valid and the
pressure drop can be taken as 2 in H2O/ft( 166.7 mm H2O/m)
∆ P flood = 0.115 F 0.7
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+ Procedure to Determine Limiting Flow Rateand Tower Diameter
1. A suitable random packing or structured packing is selected, giving an Fp value
2. A suitable GL/GG is selected along with the total gas flow rate
3. The pressure drop at flooding is calculated
4. The flow parameter is calculated, and using the pressure drop at flooding and either fig 10.6-5 or10.6-6, the capacity parameter is read off the plot
5. Using the capacity parameters, the GG is obtained, which is the maximum at flooding
6. Using a suitable % of the flooding value of GG for design, a new GG and GL are obtained. Thepressure drop can also be obtained from the plot.
7. Knowing the total gas flow rate and GG, the tower cross sectional are and ID can be calculated.
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+Example 10.6-1
Ammonia is being absorbed in a tower using pure water at 25C
and 1 atm abs pressure. The feed rate is 1440 lbm/h (653 kg/h)and contains 3 mol% ammonia in air. The process design
specifies a liquid to gas mass flaw rate ration GL/GG of 2/1 and
the use of 1-in metal Pall rings.
a) calculate the pressure drop n the packing and gas mass
velocity at flooding. Using 50% of the flooding velocity,calculate the pressure drop, gas and liquid flows and tower
diameter.
b) Repeat(a) above but use Mellapak 250Y structured
packing.
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+ Design of Plate Absorption Tower
Operating line assumption
solute A diffusing through a stagnant gas B
the moles of pure air and pure water remain constant throughout the
entire tower
similar equation to countercurrent stages process
Overall material balance
Balance around the dashed-lined box
If x and y are very dilute, (1-x) and (1-y)can be taken as 1.0 and the operating lineis straight with a slope L’/V’
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+ Number of Tray
Graphical determination of the number of trays
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+Example 10.6-2
A tray tower is to be designed to absorb SO2 from an air stream
by using pure water at 293 K. the entering gas contains 20 mol% SO2 and that leaving 2 mol % at the total pressure of 101.3
kPa . The inert air flow rate is 150 kg air/h.m2 and the entering
water flow rate is 6000 kg water/h.m2. Assuming an overall tray
efficiency of 25 %, how many theoretical trays and actual trays
are needed? Assume that the tower operate at 293 K.
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+Design of Packed Tower for Absorption
solute A diffusing through a
stagnant gas B Overall material balance
Balance around the dashed line box
give a curve line on yx plot
If x and y are very dilute, (1-x) and (1-y)can be taken as 1.0 and the operating lineis straight with a slope L’/V’
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+ Location of Operating Line
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+Limiting and Optimum L’/V’ Ratios
In absorption, V1, y1, y2, x2 is always known.
Only amount of the entering liquid flow (L2
or
L) is open to choice.
L’min – when operating line has a minimum
slope and touch (becoming tangent) to the
equilibrium line.The value of x1 is x1max.
To solve L’min, insert the value y1 and x1max in
operating line equation.
The choice of optimum L’
/V’
ratio dependon an economic balance
too high ratio value require a large liquid flow
and a large diameter tower.
too small value give small liquid flow result in a
high tower,which is also costly.
An optimum liquid flow rate can be taken as
1.2-1.5 times L’ min.
For stripping, optimum gas flow rate is takenat about 1.5 timesV’ min.
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+ Analytical Equations for TheoreticalNumber of Steps or Trays
Absorption
Procedure (for varying A):
1. Calculate m1 and m2
2. Calculate A1 and A2
3. Calculate geometric average area, A ave.= /A1A2
4. Used m2 in equation
Stripping
Procedure (for varying A):
1. Calculate m1 and m2
2. Calculate A1 and A2
3. Calculate geometric average area, A ave.= /A1A2
4. Used m1 in equation
N = In
y1 −mx2 y2 − mx2
1− 1 A
+ 1
A
InA N =
In ( x2 − y1 / m x1 − y1 / m
) 1− A( ) + A
In(1 / A)
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+Example 10.6-3
A tray tower is absorbing ethyl alcohol from an inert gas stream
using pure water at 303K and 101.3 kPa. The inlet gas streamflow rate is 100 kmol/h and it contains 2.2 mol% alcohol. It is
desired to recover 90% of the alcohol. The equilibrium
relationship is y=0.68x for this dilute stream. Using 1.5 times
the minimum liquid flow rate. determine the number of trays
needed. Do this graphically and also using the analytical
equations.
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+ Film and Overall Mass TransferCoefficients in Packed Towers
Defining a as interfacial area in m2 per m3 volume of packed
section, the volume of packing in a height dz m is S dz, where S ism2 cross sectional area of tower. Also
dA = aS dz
The volumetric film and overall mass transfer coefficient are then
defined as
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+Design Method for Packed Tower using MassTransfer Coefficient
Absorption of A from stagnant B and the moles of A leaving V equal
to the moles entering L (i.e d( Vy)= d(Lx)
Using film coefficient
Using overall coefficient
The equilibrium and operating
lines are usually curved, and k’xa,
k’ya, K’ya and K’xa vary somewhat with total gas and liquid flows
Then, these equations must be
integrated numerically or
graphically
+ Simplified Design Method for Absorption of 80
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+ Simplified Design Method for Absorption of Dilute Gas Mixture in Packed Towers
Assumptions
Concentration can be considered dilute when mole fraction y or xare less than about 0.10 (10%).
Operating line is straight
L = (L1+L2)/2
V = (V1+V2)/2
M x M y
M i x M i y
x xazK x xS
L y yazK y y
S
V
x xazk x xS
L y yazk y y
S
V
)(')()(')(
)(')()(')(
*
21
*
21
2121
+ Simplified Design Method for Absorption of 81
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+ Simplified Design Method for Absorption of Dilute Gas Mixture in Packed Towers
1. Plot operating line and equilibrium line
2. Determine interface composition yi1and xi1 at point y1, x1. To get this , findintersection point by draw a line toequilibrium line with slope
Then determine interface compositionyi2, xi2 at point y2,x2 using the samemethod. Do trial and error if necessary.
If overall mass transfer coefficient use,find appropriate y1
*, y2*, x1
* or x2*.
3. Calculate related (y-yi)M or (y-y*)M for
gas coefficient and (xi-x)M or (x*-x)M.
4. Calculate the column height byappropriate simplified equation.
)1 /('
)1 /('
1
1
yak
xak slope
y
x
+E l 10 6 4
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Example 10.6-4
Acetone is being absorbed by water in a packed tower having
a cross sectional area of 0.186 m2 at 293 K and 101.32 (1 atm).The inlet air contains 2.6 mol % acetone and outlet 0.5%. The
gas flow is 13.65 kg mol inert air/h. The pure water inlet flow is
45.36 kg mol water/h. Film coefficients for the given flows in
the tower are k’y a = 3.78 x 10-2 kg mol/s.m3.mol frac. And k’x a
= 6.16 x10-2 kg mol/s.m3.mol frac. Equilibrium data are given in
Appendix A.3.
(a) Calculate the tower height using k’y a.
(b) Repeat using k’x a.
(c) Calculate K’y a, and the tower height.
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g p
Concentrated Mixtures in Packed Tower
The design equation must be integrated graphically or
numerically.
1
2*
*
1
2*
*
1
2
1
2
))(1()1(
'))(1(
)1(
'
))(1()1(
'))(1(
)1(
'
x
x
M
x
y
y
M
y
x
x
i
iM
x
y
y
i
iM
y
x x x x
aS K
Ldx z
y y y y
aS K
Vdy z
x x x x
aS k
Ldx z
y y y y
aS k
Vdy z
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Concentrated Mixtures in Packed Tower
1. Plot operating line and equilibrium line
2. Calculate value k’ya and k’xa from empirical equation given
3. Use trial and error to determine interface composition at several
point between P1(y1,x1) and P2(y2,x2) by draw a line with following
slope from the point to intersect to equilibrium line.
4. Using the value yi and xi, calculate the value f(y) as follows. Then
plot f(y) vs. y. Calculate numerically or graphically area under
curve to insert into design equation for height.
iM y
iM x
yak
xak slope
)1 /('
)1 /('
))(1()1(
'
)(
i
iM
y y y y yaS k
V y f
+E l 10 7 1
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Example 10.7-1
A tower packed with 25.4 mm ceramic rings is to be designed
to absorb SO2 from air by using pure water at 293 K and 1.013 x105 Pa abs pressure. The entering gas contains 20 mol % SO2and that leaving 2mol %. The inert air flow is 6.53 x10-4 kg mol
air /s. the tower cross-sectional area is 0.0929 m2. For dilute SO2,the film mass transfer coefficients at 293 K are for 25.4 mm
rings,
k’ya = 0.0594 Gy0.7 Gx
0.25 and k’xa = 0.152GX0.82.
where k’ya & k’xa is kg mol/s.m3. mol frac., and Gx ,Gy are kg
total liquid or gas, respectively, per sec per m2
tower crosssection. Calculate the tower height.
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+ Design Of Packed Towers Using Transfer Units: 87
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Design for Dilute Solutions
The terms (1-y)iM/(1-y), (1-x)iM/(1-x), (1-y)*M/(1-y) and (1-x)*M/(1-x)
are close to one, and if the operating lines and equilibrium line arestraight and dilute,
z = H G N G = H Gdy
y− yi y2 y1∫ = H G × y1 − y2
( y − yi ) M
z = H OG N OG = H OG dy y− y* y2 y1
∫ = H G × y1 − y2( y − y* ) M
z = H L N L = H Ldx
xi − x x2 x1∫ = H L × x1 − x2
( xi − x) M
z = H OL N OL = H OLdx
x* − x x2 x1
∫ = H OL ×
x1 − x2( x* − x) M
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Design for Dilute Solutions
Using the equilibrium line equation y = mx and letting A = L/mV, the number
of transfer unit for absorption and stripping are
When the operating and equilibrium lines are straight and not parallel, NOGis related to the number of theoretical trays N by equation:
The height of theoretical trays (HETP)
N OG = 1
(1−1 / A) In[(1−1 / A)( y1 −mx2
y2 −mx2) +1 / A]
N OL = 1
(1− A) In[(1− A)( x2 − y1 / m
x1 − y1 / m) + A]
HETP = H OG In(1 / A)
(1− A) / AThen, z = N x HETP
N OG = N In A
(1−1 / A)
+E l 10 6 5
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Example 10.6-5
Acetone is being absorbed by water in a packed tower having
a cross sectional area of 0.186 m2 at 293 K and 101.32 (1 atm).The inlet air contains 2.6 mol % acetone and outlet 0.5%. The
gas flow is 13.65 kg mol inert air/h. The pure water inlet flow is
45.36 kg mol water/h. Film coefficients for the given flows in
the tower are k’y a = 3.78 x 10-2 kg mol/s.m3.mol frac. And k’x a
= 6.16 x10-2 kg mol/s.m3.mol frac. Equilibrium data are given in
Appendix A.3.
a) Use HG and NG to calculate tower height
b) Use HOG and NOG to calculate tower height
c) Use E 10.6-52 to calculate NOG and tower height
d) Using analytical equations, calculate HETP, theoreticalsteps N and tower height.