Gas Absorption

Absorption Reading: Chap. 13

• • • •

Definition Equipment Packing materials Design considerations: – Mass balance – High gas flow – Mass flow

• Concentrated systems • HTU and NTU

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Definition Transfer 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 it’s gas phase or liquid phase controlled? 11/21/08

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Equipment Spray tower

Clean gas out Countercurrent Clean gas out

Spray nozzle

Dirty gas in

Q: Limitations of a spray tower? 11/21/08

Redistributor Q: Why redistributor? Aerosol & Particulate Research Lab

packed tower Mist Eliminator Liquid Spray Packing

Dirty gas in

Liquid outlet Mycock et al., 1995 3

Three-bed cross flow packed tower

Liquid spray

Dry Cell

Packing

Berl Saddle

Intalox Saddle

Raschig Ring

Lessing Ring

Q: Criteria for good packing materials? 11/21/08

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Pall Ring

Tellerette Mycock et al., 1995 4

Design considerations: What are known? What are we looking for? 11/21/08

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Mass Balance

Gas out

In = Out Liquid in

Gas in

Liquid out

Gm1 + Lm 2 = Gm 2 + Lm1 Gm ( 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 gas 11/21/08

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Slope of Operating Line = Lm/Gm

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Dirty air

Clean air

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Clean water

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Dirty water 7

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. Henry’s 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

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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: 1.17×10-3 g/cm3. Properties of water:: density: 1 g/cm3; viscosity: 0.8 cp. 11/21/08

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(G ' ) 2 FΦµ 0L.1 ρG ρ L g L: 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 Φ: specific gravity of the scrubbing liquid µL: liquid viscosity (in cP; 0.8 for water) 11/21/08

L G

ρG (dimensionless) ρL

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Mass Transfer rate of mass Flux =   transferred

 /  interfacial  = k  concentration    area   difference      

J ( = M / A) = k ( Ci − C ) mass ) area × time

J: flux k: mass transfer coefficient (

Two-Film Theory (microscopic view)

CI

J = k G ( pG − pI )

CL

(gas phase flux)

J =k L ( C I − C L )

pG

(liquid phase flux)

pI = HC I

pI

1 ( pG − HCL ) J= 1 / kG + H / k L Cussler, “Diffusion”, Cambridge U. Press, 1991.

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(overall flux)

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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* = HC L 1 / k + H / k (overall gas phase MT coefficient) G L (equivalent pressure to the 2

bulk 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

1 11/21/08

G 'm ⇒ x = x1 + ( y − y1 ) L 'm Aerosol & Particulate Research Lab

L’m: molar flux of liquid G’m: 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 y ( y − y *) K aP OG (tower height) 1

Z

a: packing area per volume

y* = Hx

 y1 − Hx1  1  ⇒Z = ln K OG aP (1 / G 'm − H / L'm )  y Z − HxZ   y1 − Hx1  G 'm 1  = × ln K OG aP (1 − HG 'm / L'm )  y Z − HxZ  1

HTU? 11/21/08

NTU? Aerosol & Particulate Research Lab

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Mass balance

x1, y1

L'm ( x − x1 ) y = y1 + G 'm Equilibrium

y* = Hx

x 1, y 1*

y1 G 'm dy Z= K OG aP ∫y Z ( y − y *)

xZ, yZ

xZ, yZ*

Alternative solution:

G 'm y1 − y z Z= × ; K OG aP ∆y LM

∆y LM

( y − y )−(y =

* − y 1 z z  y1 − y1*   ln *   yz − yz  * 1

)

Assumptions for dilute/soluble systems? 11/21/08

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Pure amine Lm = 0.46 gmole/s

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 tower’s cross-sectional area is 0.84 m2. KOGa = 9.34×10-6 s-1atm-1cm-3. The pressure is 1 atm. Determine the tower height that can achieve this goal.

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0.04% CO2

1.27% CO2 Gm = 2.31 gmole/s C* = 7.3% CO2 in amine

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Absorption of concentrated vapor Mole balance on the controlled volume

x 1, y 1

d d 0 = − (G 'm y ) + ( L'm x) dz dz Gas flux

 1   G 'm = G 'm 0  1− y 

Liquid flux  1  L 'm = L ' m 0   1− x 

x1 , y1 *

xZ, yZ

 y1  L'm 0  x x1    +   − 1 − y1  G 'm 0  1 − x 1 − x1   ⇒y=  y1  L'm 0  x x1   +   1 +  −  1 − y1  G 'm 0  1 − x 1 − x1  11/21/08

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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 ) dz ⇒Z =∫

Z

0

G 'm 0 y1 dy dz = = HTU × NTU 2 ∫ y K OG aP Z (1 − y ) ( y − y *)

G 'm0 HTU = K OG aP

NTU = ∫

y1

yZ

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dy 2 (1 − y ) ( y − y*)

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HTU (ft)

HTU

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For a given packing material and pollutant, HTU does not change much. Aerosol & Particulate Research Lab

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Quick Reflection

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