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A good presentatin for gas absorption review.
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AbsorptionReading: Chap. 13
Definition Equipment Packing materials Design considerations: Mass balance High gas flow Mass flow
Concentrated systems HTU and NTU
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11/21/08
Aerosol & Particulate Research Lab
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DefinitionTransfer of a gaseous component (absorbate) from the gas phase to a liquid (absorbent) phase through a gas-liquid interface.Q: What are the key parameters that affect the effectiveness? Q: How can we improve absorption efficiency?
Mass transfer rate:
gas phase controlled absorption liquid phase controlled absorption
Q: Does it matter if its gas phase or liquid phase controlled?11/21/08 Aerosol & Particulate Research Lab 2
EquipmentSpray towerClean gas out Clean gas out Countercurrent
packed towerMist Eliminator Liquid Spray Packing
Spray nozzle
Dirty gas in
Dirty gas in
Q: Limitations of a spray tower?11/21/08
Redistributor Q: Why redistributor?Aerosol & Particulate Research Lab
Liquid outletMycock et al., 1995 3
Three-bed cross flow packed tower
Liquid spray
Dry Cell
Packing
Berl Saddle11/21/08
Intalox Saddle
Raschig Ring
Lessing Ring
Pall Ring
TelleretteMycock et al., 1995 4
Q: Criteria for good packing materials?Aerosol & Particulate Research Lab
Design considerations: What are known? What are we looking for?11/21/08 Aerosol & Particulate Research Lab 5
Mass BalanceIn = OutLiquid in
Gas out
Gas in
Liquid out
Gm1 + Lm 2 = Gm 2 + Lm1Gm ( y1 y2 ) = Lm ( x1 x2 )(for a dilute system)
Lm: molar liquid flow rate Gm: molar gas flow rate x: mole fraction of solute in pure liquid y: mole fraction of solute in inert gas11/21/08 Aerosol & Particulate Research Lab
Slope of Operating Line = Lm/Gm
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Dirty air
Clean air Clean water Dirty water7
11/21/08
Aerosol & Particulate Research Lab
Generally, actual liquid flow rates are specified at 25 to 100% greater than the required minimum.
Q: How much is X2 if fresh water is used? What if a fraction of water is recycled?
G = 84.9 m3/min (= 3538 mole/min). Pure water is used to remove SO2 gas. The inlet gas contains 3% SO2 by volume. Henrys law constant is 42.7 (mole fraction of SO2 in air/mole fraction of SO2 in water). Determine the minimum water flow rate (in kg/min) to achieve 90% removal efficiency.11/21/08 Aerosol & Particulate Research Lab 8
Problems with high gas flow Channeling: the gas or liquid flow is much greater at some points than at others Loading: the liquid flow is reduced due to the increased gas flow; liquid is held in the void space between packing Flooding: the liquid stops flowing altogether and collects in the top of the column due to very high gas flow Gas flow rate is 3538 mole/min and the minimum liquid flow rate is 2448 kg/min to remove SO2 gas. The operating liquid rate is 50% more than the minimum. The packing material selected is 2 ceramic Intalox Saddles. Find the tower diameter and pressure drop based on 75% of flooding velocity for the gas velocity. Properties of air:: molecular weight: 29 g/mole; density: 1.1710-3 g/cm3. Properties of water:: density: 1 g/cm3; viscosity: 0.8 cp.11/21/08 Aerosol & Particulate Research Lab 9
(G ' ) 2 F 0.1 L G L gL: mass flow rate of liquid G: mass flow rate of gas G: mass flux of gas per cross sectional area of column F: Packing factor : specific gravity of the scrubbing liquid L: liquid viscosity (in cP; 0.8 for water)11/21/08
L G
G (dimensionless) L10
Aerosol & Particulate Research Lab
Mass Transfer J ( = M / A ) = k ( Ci C )J: flux k: mass transfer coefficient( mass ) area time
rate of mass Flux = transferred
/ interfacial = k concentration area difference
Two-Film Theory (microscopic view)
CI
J = k G ( pG pI )(gas phase flux) (liquid phase flux)
J =k L ( C I C L )
CL pG pI
pI = HC I
1 ( pG HCL ) J= 1 / kG + H / k L Cussler, Diffusion, Cambridge U. Press, 1991.(overall flux)Aerosol & Particulate Research Lab 11
11/21/08
1 pG K OL = C* = J = K OL ( C* C L ) H (overall liquid phase MT coefficient) 1 / k L + 1 / k G H (equivalent concentration 1 to the bulk gas pressure) = K OG ( pG p* ) K OG = p* 1 / kG + H / k L (equivalent = HC L the (overall gas phase MT coefficient) pressure to2bulk concentration in liquid)
Macroscopic analysis of a packed tower Mole balance on the solute over the differential volume of tower
accumulation = flow of solute in of solute minus flow out
dy dx 0 = G 'm + L 'm dz dz
111/21/08
G 'm x = x1 + ( y y1 ) L 'mAerosol & Particulate Research Lab
Lm: molar flux of liquid Gm: molar flux of gas
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Mole balance on the solute in the gas only
solute = solute flow in solute lost accumulation minus flow out by absorption
dy 0 = G 'm K OG aP ( y y*) dz Z y G 'm dy Z = dz = 0 K OG aP y ( y y *) (tower height)1 Z
a: packing area per volume
y* = Hx
y1 Hx1 1 Z = ln y Hx K OG aP (1 / G 'm H / L'm ) Z Z y1 Hx1 G 'm 1 = ln y Hx K OG aP (1 HG 'm / L'm ) Z Z 1HTU? NTU?Aerosol & Particulate Research Lab 13
11/21/08
L'm ( x x1 ) y = y1 + G 'mEquilibrium
Mass balance
x1, y1
y* = Hxy1 G 'm dy Z= K OG aP y Z ( y y *)
x 1, y 1* xZ, yZ
Alternative solution:
xZ, yZ*
G 'm y1 y z Z= ; K OG aP y LM11/21/08
y LM
(y y )(y =* 1
Assumptions for dilute/soluble systems?
y* 1 z z * y1 y1 ln y y* z z
)14
Aerosol & Particulate Research Lab
Pure amine Lm = 0.46 gmole/s
0.04% CO2
Q: A Packed tower using organic amine at 14 oC to absorb CO2. The entering gas contains 1.27% CO2 and is in equilibrium with a solution of amine containing 7.3% mole CO2. The gas leaves containing 0.04% CO2. The amine, flowing counter-currently, enters pure. Gas flow rate is 2.31 gmole/s and liquid flow rate is 0.46 gmole/s. The towers cross-sectional area is 0.84 m2. KOGa = 9.3410-6 s-1atm-1cm-3. The pressure is 1 atm. Determine the tower height that can achieve this goal.
1.27% CO2 Gm = 2.31 gmole/s C* = 7.3% CO2 in amine
11/21/08
Aerosol & Particulate Research Lab
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Absorption of concentrated vaporMole balance on the controlled volume
d d 0 = (G 'm y ) + ( L'm x) dz dz 1 G 'm = G 'm 0 1 y
x 1, y 1x1 , y1 *
Gas flux
Liquid flux 1 L 'm = L ' m 0 1 x
xZ, yZ
y1 L'm 0 x x1 1 y + G' 1 x 1 x 1 m0 1 y= y1 L'm 0 x x1 1+ 1 y + G' 1 x 1 x 1 m0 1 11/21/08 Aerosol & Particulate Research Lab
xZ, yZ*
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Mole balance on the gas in a differential tower volume
G 'm 0 dy 0= K OG aP ( y y*) 2 (1 y ) dzZ =Z 0
G 'm 0 y1 dy dz = yZ (1 y) 2 ( y y *) = HTU NTU K OG aP
G 'm0 HTU = K OG aP
NTU =
y1
yZ
dy 2 (1 y ) ( y y*)
11/21/08
Aerosol & Particulate Research Lab
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HTU (ft)
HTU
11/21/08
For a given packing material and pollutant, HTU does not change much. Aerosol & Particulate Research Lab
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Quick Reflection
11/21/08
Aerosol & Particulate Research Lab
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