absorption of concentrated solutions · pdf filechapter 6: rate-based absorption 56 predicting...

Chapter 6: Rate-based Absorption 46

Absorption of Concentrated Solutions

Last lecture:

• Extended our analysis of absorption as a separation technique to include the use of two film theory combined with material balances for analyzing packed bed absorption processes for dilute solutions.

This lecture will focus on:

• Extending our analysis of absorption to include the use of two film theory for analyzing packed bed absorption processes for concentrated solutions.

• Discussing how to predict the overall and the film mass transfer coefficients.


Gas Phase Material Balance

• Solute A mole balance

( ) ( ) ( )in AG ,in out AG,out interface A cross AV y V y A N a A z NΔ− = =

Vout yAG,out

Vin yAG,in

Δz

Across

( ) ( )( ) ( )y *

AG AG A crossA LM

K ad V y y y A dz

1 y−

= −−

( ) ( )y *A AG A

A LM

KN y y

1 y= −

−

• Flux in terms of the overall mass transfer coefficient :

• Limit as Δz → 0 :

( ) ( ) ( )*A AG

A LM *A

AG

1 y 1 y1 y

1 yLn1 y

− − −− =

⎡ ⎤−⎢ ⎥−⎣ ⎦

Lin,xin

Vout,yout

Vin,yinLout,xout

Z

Δz

Across


Solving the Material Balance Equation

• Combine thelast 2 equations : ( )

( )( ) ( )y *AG

AG A crossAG A LM

K ad yV y y A dz1 y 1 y

−= −

− −

• Separate variables& integrate ( 1 refers to inlet, while 2 refers to outlet) :

( )( ) ( )( )

( ) ( )( )

1

2

1

2

yzAG

col0 y y cross *

AG AG AA LM

yAG

*y y cross AG AG A

V d yl dzK a A

1 y y y1 y

V d yK a A 1 y y y

= =

− −−

=′ − −

∫ ∫

∫

( )( ) ( )

AG AG AGAG AG 2

AG AG AGAG

y d y d yVd V y d y V d V V1 y 1 y 1 y1 y

⎛ ⎞ ⎛ ⎞′′ ′= = = =⎜ ⎟ ⎜ ⎟− − −−⎝ ⎠ ⎝ ⎠

• A little calculus magic!! (V’ is solute free flow-rate) :

( ) ( )( ) ( )y *

AG AG A crossA LM

K ad V y y y A dz

1 y−

= −−

The problem is that when the gas phase is concentrated and the solute MT rate between phases is large…NONE of the variables inside the integral except (hopefully) the cross-sectional area of the column remains constant from the bottom of the column to the top of the column !


Material Balance on a Packed Column

• General solution: ( )( ) ( )( )

1

2

yAG

coly y cross *

cross AG AG AA LM

V d ylK a A

A 1 y y y1 y

=

− −−

∫

• Use average values of V, Kya and assume Across is constant

( )( )

( )( )1

2

yA LM

col AG*yy cross AG AG A

1 yVl d yK a A 1 y y y

−=

− −∫

• Definitions :

( )OGy cross

VHK a A

=

( )( )( )

1

2

yA LM

OG AG*y AG AG A

1 yN d y

1 y y y−

=− −∫

col OG OGl H N=

HOG = height of a transfer unit based on the overall gas phase MT coefficient

NOG = number of transfer units required to accomplish a given separation based on the overall gas phase MT coefficient

yA is the gas phase mole fraction of solute A evaluated at each end of the column. This term appears in (1-yA)LM

yAG is the gas phase mole fraction of solute A evaluated at any position within the column. This is the variable of integration in determining NOG.


Dilute Solutions

• Everywhere in the column :

( )( )( )

1

2

yA LM

OG AG*y AG AG A

1 yN d y

1 y y y−

=− −∫

A1 y 1− ≅

• Therefore :

( )( )

A LM

AG

1 y1

1 y−

≅−

( )1

2

yAG

OG *y AG A

d yNy y

=−∫

• Similar equations apply for the other types of transfer units…NOL, NG & NL

( )1 2

OG *AG A LM

y yNy y

−=

−

• If both the operating curve and the equilibrium curve are linear, then…


Use Mole Ratios to Linearize Operating Curve

• If we use X & Y instead of x and y, the operating curve will be linear.

– This can make it easier to evaluate this integral

– Especially if you fit the equilibrium data to a polynomial and evaluate the integral in Excel

– For dilute solutions, this yields:

( )1

2

yAG

OG *y AG A

d yNy y

=−∫

( )1

2

YAG

OG *Y AG A

d YNY Y

=−∫

OGy cross

VHK a A

′=

A AA A

A A

Y yy or Y1 Y 1 y= =+ −


Concentrated Absorption Using HOG NOG

• General solution :( )OG

y cross

VHK a A

=

( )( )( )

1

2

yA LM

OG AG*y AG AG A

1 yN dy

1 y y y

−=

− −∫

col OG OGl H N=

• Concentrated solutions, use mole ratios– Linearize the operating curve

( )OGy cross

VHK a A

′=

( )( )( )

1

2

YA LM

OG AG*Y AG AG A

1 YN dY

1 Y Y Y

−=

− −∫

col OG OGl H N=

• HOG is still easy to evaluate

• You must numerically integrate to determine NOG.

– You must keep ALL the terms !


Concentrated Absorption Using k′xa & k′ya

• Last lecture :

( )Gy cross

VHk a A

=

( )( )( )

1

2

yA LM

G AGAG AG Aiy

1 yN dy

1 y y y−

=− −∫

col G Gl H N=

• Concentrated solutions, use mole ratios– Linearize the operating curve

• HG is still easy to evaluate

• You must numerically integrate to to determine NG.

– You must keep ALL the terms !

( )Gy cross

VHk a A

′=

( )( )( )

1

2

YA LM

G AGAG AG AiY

1 YN dY

1 Y Y Y−

=− −∫

col G Gl H N=


Alternative Mass Transfer Coefficient Groupings


Predicting Overall Mass Transfer Coefficients

• The relationship between film and overall MT coefficients is :

y y x

1 1 mK a k a k a

= +

• Where m is the slope of the equilibrium curve

• Next use the definitions of HG, HL and HOG

( )Gy cross

VHk a A

=( )L

x cross

LHk a A

= ( )OGy cross

VHK a A

=

• Combining yields :

LOG G

HH H

A= + OL L GH H A H= +

LAmV

=

• You must evaluate m, L, V, A, HG, HL, HOL and HOG at the top of the column and at the bottom of the column and use the average of the HOG or HOL values that you obtain


Predicting Film Mass Transfer Coefficients

• Data must be collected and correlated

– Even these do not yield very accurate predictions

– Problem is that it is not easy to measure film coefficients

• Try to get a system in which either gas-side or liquid-side resistance to MT is negligible.

• Cannot predict MT coefficient values from theory

– Problem is too complex.

– Depend on liquid and gas properties, transport properties, fluid flow phenomena and packing material characteristics



• The Correlations

– When gas-phase MT resistance is dominant, use HG

– Where

• The constants CV, ε and a can be found in Table 6.8 (see next slides)

• hL is the liquid hold-up in the column

• Revap is the Reynolds Number for the vapor

• Scvap is the Schmidt Number for the vapor

• uvap is the superficial vapor velocity

• Dvap is the diffusivity of the solute in the vapor phase

• aPh / a is the ratio of the phase interface area to the surface area of the packing material

( )1 / 2

1 / 2 vap3 / 4 1 / 3G y L vap vap4

V vap Ph

u a1 4H H h Re ScC D aa

εε − −⎛ ⎞⎛ ⎞= = − ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠



• The Correlations– When liquid-phase MT resistance is dominant, use HL

– Where

• The constants CL, ε and a can be found in Table 6.8 (see next slides)

• hL is the liquid hold-up in the column

• uliq is the superficial liquid velocity

• Dliq is the diffusivity of the solute in the liquid phase

• aPh / a is the ratio of the phase interface area to the surface area of the packing material

1 / 21 / 6liqL

L xL liq liq Ph

u4 h1 1 aH HC 12 D au a a

ε⎛ ⎞ ⎛ ⎞⎛ ⎞= = ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠


A Million Additional Equations

( )( )

vap vapvapvap

vap vap vap

MomentumdiffusivitySc

D Mass diffusivity D

μ ρμρ

= =

1 / 3 2 / 3liq h

Lliq

12 Fr ah

Re a⎛ ⎞ ⎛ ⎞= ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

liq liqliq

liq

uRe

aρμ

=

2liq

liq

u a inertial forceFr Froudenumberg gravitational force

= = =

0.15 0.1hh liq liq liq

aC Re Fr for : Re 5

a= <0.25 0.1h

h liq liq liqa

0.85 C Re Fr for : Re 5a= ≥


Heat Effects & Heat of Solution

• When some solutes, especially strong acids or bases are absorbed into a liquid, the heat of solution causes the temperature of the solution to increase

• As a result…

– The temperature of the liquid can be greater than the temperature of the gas

– The temperature of the liquid is greater near the bottom of the column

• This problem is complicated and generally requires computational assistance (Aspen…)

• Simplified Adiabatic Design Method

– Simple, approximate method

– Helps you assess the validity of results obtained from Apsen

– Takes heat of solution into account

– Approximate effect of liquid temperature variation on the equilibrium curve is included in this method


Homework

First: Practice through solving Examples (6.12, 6.14, and 6.15)

Homework

Problem 6.32 Problem 6.33Problem 6.35 Problem 6.37

Problem 6.39

Due Date: 10th April, 2007

absorption of concentrated solutions · pdf filechapter 6: rate-based absorption 56 predicting...

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