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Department of Chemical & Biomolecular Engineering

THE NATIONAL UNIVERSITYof SINGAPORE

Chemical Engineering Process Laboratory II

Semester 5

Experiment R1

Chemical Absorption in Packed Bed

Yeo Kar Ling Catrina U046901B

Yu Haoli Henry U046865B

Zhang YanYang U046709X

Zhou Yien U046854U

Group : F15 Date of Expt. : 01 September 2006

Demonsta tors Signature:

Grade :

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Summary

In this experiment, the objective is to investigate the various reaction zones in a

continuous, counter-current, gas-liquid absorption with chemical reaction in a packed bed.

We design an experiment to determine the heights of the various reaction zones as well as

the mass transfer coefficients in a packed column for the absorption of CO 2 in an aqueous

solution of NaOH.

A total of 6 experimental runs were conducted and the compositions of both inlet and

outlet gas streams were measured and recorded. The volumetric flow rates of both liquid

and the gaseous inlet streams were kept constant. The concentration in the liquid stream

was gradually increased so as to analyse the effect of chemical reaction between CO 2 gas

and aqueous NaOH on the absorption of gaseous CO 2.The overall mass transfer coefficients for both the liquid and gaseous phases were then

determined and the heights of the physical absorption, interior reaction and surface

reaction zones determined. The composition (mole ratio of CO 32- and OH -) of the liquid

outlet was determined by double titration using first, phenolphthalein and then methyl

orange. The inlet and outlet gas composition was determined by gas chromatography.

The enhancement factor E for each of the 6 runs was calculated to compare the

effectiveness of the chemical reaction in the absorption zones. These results are presented

in the table on page 28.

From the analysis of the data obtained, it was observed that as the concentration of NaOH

solution in the liquid stream was increased, the absorption of CO 2 from the gaseous

stream also increased. Hence, the presence of chemical reaction between NaOH and CO 2

has enhanced the efficiency of absorption of CO 2. The extent of improvement in the

efficiency of absorption can be seen from the enhancement factor, E which increases as

amount of NaOH present in the liquid stream increases.

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Table of Contents

I. Introduction ............. 4

II. Technical / Theoretical Background. 5

III. Experimental . .. 14

IV. Results and Calculation . ..... 18

V. Discussion .................................... .... 32

VI. Error Analysis 38

VII. Conclusion .. 39

VIII. References .. 40

IX. Appendix A. .. . 41

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Introduction

Many commercial applications today make use of gas absorption processes involving

systems which take place in the liquid phase. These range from the manufacture of paper

to organic chemicals manufacture. They are also present in pollution control systems likescrubbers in gas furnace exhaust towers. The reaction of NaOH with CO 2 enhances the

rate of absorption and increases the capacity of the liquid solution to dissolve the solute

Packed bed gas-liquid absorption operations are frequently improved by utilizing a

chemical reaction to reduce mass transfer resistance. This is generally effective for fast or

instantaneous reactions in the liquid phase. Similarly, the absorption of CO 2 gas can be

improved by introducing aqueous NaOH solution into the system thus allowing the

following chemical reaction to take place to speed up the absorption.

O H CO Na NaOH CO 2322 2

In a continuous, counter current packed bed absorption process with chemical reaction.

There can be three distinct absorption zones, depending on the operating conditions.

They are the surface reaction zone, interior reaction zone and the physical absorption

zone.

The aim of this experiment is to study the various zones in a continuous, counter current,

gas absorption with chemical reaction in a packed bed as well as the mass transfer

coefficients in a packed bed for the absorption of gaseous CO 2 in an aqueous solution of

NaOH.

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Technical Background (Theoretical Background)

Packing towers are extensively used in chemical industries for gas absorptions.

Gas absorption is a mass transfer operation that involves a soluble component being

transferred from gas medium to a liquid medium that it is soluble by contact. In a packing

tower, gas stream and liquid stream usually will flow counter-currently. To achieve

desirable recovery of solute from the gas phase, the liquid carrier should be relatively

involatile and the gas carrier should be relatively inert to the liquid. In many cases, in

order to increase the efficiency of absorption, chemical reactions will be involved to

increase the recovery of solute from gas stream by the liquid stream. A liquid will be

chosen such that it will react with the desired component in the gas medium but not the

other components in the air.

The most commonly used method of describing interfacial mass transfer is the

two-film theory of Whitman. Here the mass transfer is assumed to take place only in two

stagnant films on either side of the interfacial area. Assuming Ficks law, the mass

transfer coefficient is given by the binary diffusion coefficient and the film thickness.

The two-film theory of Lewis and Whitman defines the resistance of mass

transfer from gas phase to liquid phase as a gas film resistance and a liquid film

resistance. The resistances of the two films can be combined in an overall resistance. The

concentration gradient between the material in the bulk liquid or gas and the material in

the liquid or gas at the interface is considered as the driving force of mass transfer.. In

practice, it is quite impossible to measure interfacial conditions and overall coefficients

are used giving the equation

dw/dt = K LA (x* - x) = KgA(y - y*)

where dw/dt is the amountof gas passing through the interface per unit time, K L is the

overall liquid mass-transfer coefficient, Kg is the overall gas mass-transfer coefficient, A

is the interfacial area and x, y are the concentrations of the gas being transferred, in the

liquid and gas streams respectively. The quantities of x* and y* are introduced into the

equation because usually concentrations in the liquid and in the gas are expressed in

different units ( e.g. mol/l for concentration in liquid while partial pressure for gas

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concentration). x* represents the concentration in the liquid which would be in

equilibrium with a concentration y in the gas and y* the concentration in the gas stream

which would be in equilibrium with a concentration x in the liquid.

The overall mass transfer resistance can be found out from the following equation:

where k l = liquid-phase mass transfer coefficient (L/T)

k g = gas-phase mass transfer coefficient (L/T)

H is the Henry s constant.

k l and k g are calculated by using the following equations based on two-film theory:

Where l and g correspond to liquid and gas-phase film thickness, respectively .

In this experiment, CO 2 is the desired component to be removed from the gas

stream. Water is used to absorb CO 2. Carbon dioxide will dissolve in water to form

carbonic acid at equilibrium.

CO 2(aq ) + H 2O( l) H 2CO 3(aq ) (1)

The solubility of CO 2 in water is relatively low. At room temperature, thesolubility of carbon dioxide is about 90 cm 3 of CO 2 per 100 mL of water. Therefore, to

enhance the absorption, NaOH is used in this experiment. NaOH will react with dissolved

carbon dioxide to form sodium carbonate.

H2CO3 (aq) + NaOH (aq) NaCO3 (aq) + H 2O (l) (2)

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According to Le Chateliers Principle, the equ ilibrium (1) will shift to the right. More

CO 2 can then be absorbed from the gas stream. The reaction between CO 2 and NaOH can

be considered to be instantaneous.

The overall improvement in absorption due to chemical reaction is measured using the

enhancement factor, E.

LA

LA

k

k E

For instantaneous reaction,

Ai A

B B

A A

A B B

C bDC D

pbD H C D E 11

Inside a packing tower, 3 zones of absorption can be identified in the column depending

on the concentration of the reactants in both phases. In a counter-current operation, the

zones are shown in the following figure:

xB,in yA,out

xB3 = x BC yA3

xB2 = 0 yA2

ha

h2

h3

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Fig. 1 three absorption zones in packing zone

yA,in xB,out = 0

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Gasfilm

Liquidfilm

ReactionplanePA

CB

PAi

CAi

Zone 1: Surface reaction zone

In this zone, the gas reactant concentration is low whilst the liquid reactant

concentration is high. Reaction is at the interface. Mass transfer rate in gas phase controls

the overall rate of mass transfer.

Zone 2: Interior reaction zone

Gas reactant concentration will be higher than the liquid reactant concentration in this

zone. Reaction plane will be in liquid film. Mass transfer rates in both gas and liquid

phases will control the overall transfer rate.

Zone 3: Physical absorption zone

There will not be any chemical reaction in this zone because the liquid concentration is

zero. Only physical absorption takes place. Gas reactant concentration is high.

Gasfilm

Liquidfilm

Reactionplane

PA

HighCB

End of surfacereaction zone

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Designing packing tower:

Assume reaction system: A + bB = products

Operating line of packing tower is derived based on material balance equation of

component A.

G(Y Ain YA out ) = L(X Bin XB out )/b

G = gas carrier molar flow rate, mol/s

L = liquid carrier molar flow rate, mol/s

YA = molar ratio of A in gas phase

XB = molar ratio of B in liquid phaseTreating packing tower as a plug flow reactor, for PFTR:

A

A

x

x AV

A AO r

dxF V

0

or A

A

x

x AS

A AO r

dxF S

0

(1)

A x A AO dY GAdxF (2)

where F AO = molar feed rate of A

xA = fractional conversion of A

Ax = the cross-sectional area of the reactorSubstituting (2) into (1)

A

Ao

Y

Y AV

A x r

dY GAV or

A

Ao

Y

Y AV

A x x r

dY GAhA or

A

Ao

Y

Y AV

A

r

dY Gh

Gasfilm

Liquidfilm

PA

CAi

PAi

CAb

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A

Ao

Y

Y AS

A x r

dY GAS or

A

Ao

Y

Y AS

A x x r

dY

S V

GAhAV ie A

Ao

Y

Y AS

A

r

dY

aG

h

Zone 1: Surface reaction zone (height h 1)

Reaction plane at the gas-liquid interface, p Ai = Y Aipinert = 0

Ainert gAT A

AgA AgA

A AS Y pk PY

Y k Pk

dt

dN

S r

1

1

if P inert + P A = P T, Ainert

A Y P

P

Substituting the equation of rate into the equation for height derived above,

A

Ao

Y

Y AS

A

r

dY

aG

h3 = A

Ao

Y

Y A

A A

T gA Y

dY Y

aPk G )1(

= )]([ln 434

3 A A

A

A

T gA

Y Y Y

Y

aPk G

At the end of surface reaction zone and starting of the interior reaction zone, at the

reaction plane, Y Ai = C Bi = 0, molar ratio of B in the liquid = X BC

Based on the mass transfer equation, assuming dilute solution, (when mole fraction of the

reactant in its phase is less than 5%), X BC is given the the following equation:

1

4241

1 A B

BC

Y X X

where

o LA

gA

B

A

T

T

k

k

D D

C P

G L

1 , 12 LbG

b = stoichiometric coefficient of the liquid reactantk gA = gas mass transfer coefficient of AkgA = gas mass transfer coefficient of A without chemical reaction

XBC is the minimum concentration of B required in the liquid stream for surface reaction

to take place.

If X Bout < X BC, interior reaction zone will be present.

If X Bin < X BC, surface reactionzone will not be present.

Zone 2: Interior reaction zone (height h 2)

In interior reaction zone, X Bout = X B2 = 0.

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Overall rate equation for this region is the Hatta equation

If Y A

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However, not all the three zones will be present in the column. It will depend on the inletconcentration of reactant B in liquid phase.

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Experimental

Apparatus:

1. A column packed with Raschig Ring

Fig 1: Packed column (Raschig Ring packing)

2. Titration apparatus

(a) Burettes

(b) Pipettes

(c) Conical flask

(d) Beakers

3. A Gas Chromatograph with gas syringes

Rotameters for inflow rates

LiquidFeed Tank containing0.1MNaOH

RaschigRingPacking

Outlet as

Inlet as

ControlValve forstep 5

LiquidInlet

Liquidoutlet

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Reagents Provided:

1. 2M NaOH

2. 0.1022M NaOH as standard solution for titration only

3. 0.1097M HCL as standard solution for titration only

4. Indicators for titration: phenolphthalein and methyl orange

5. Air and CO 2 supply

Experimental Procedure

(a) Getting Started

1. About 20L of approximately 0.1M NaOH was prepared, in the feed tank, by

diluting the 2.0M NaOH given using the dilution ratio of 1:19.

2. Inflow rate for CO 2 was fixed at 0.2L/min while that for air was fixed at

1.00L/min. The inflow rates were adjusted by rotameter for the CO 2 gas stream

and air stream respectively.

3. Inflow rates for NaOH and water were adjusted similarly using the rotameters.

Inflow rate for NaOH was adjusted to zero for the first run while that for water

was adjusted to 0.6L/min.

4. The resulting system was left unperturbed for approximately 15 to 20 min to

reach steady state.

5. The control valve for the outlet bottom product was adjusted as and when

necessary to maintain the bottom water level below the gas outlet.

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(b) Collection of samples

6. After the system attained steady state, a gas sample was collected from the inlet

gas stream using a gas syringe and the sample was analyzed using the gas

chromatograph.

7. 2 gas samples from the outlet gas tube were also collected and analyzed. If the

two readings were consistent, the experiment proceeded accordingly. Otherwise,

the system was given another 5 to 10 min to reach steady state before collecting

another 2 gas samples.

8. 150ml of outlet liquid was collected as sample for titration.(c) Titration

9. 25.0ml of liquid sample was pipette and titrated against standardized HCL using

phenolphthalein as indicator to obtain the colorless end point.

10. The resulting mixture from step 9 was further titrated against the same HCL

using methyl orange as the new indicator to obtain another orange end point

(double titration).

11. Titration was carried out again using another 25.0ml of the liquid sample to

ensure consistency and accuracy in the results.

12. Steps 2 to 11 were repeated five times, using different inflow rates for water and

NaOH, i.e. inflow rates for NaOH and water in step 3 was changed accordingly

using the table below.

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Inlet Flow Rates in L/min

Run H 2O NaOH Air CO 2 1 0.60 0.00 1.0 0.22 0.55 0.05 1.0 0.23 0.50 0.10 1.0 0.24 0.45 0.15 1.0 0.25 0.40 0.20 1.0 0.26 0.35 0.25 1.0 0.2

Table 1: Six runs, with varying NaOH inflow rates, were conducted in the experiment

Precautions

1. HCL solutions are corrosive and NaOH solutions are caustic. Handle with care! If

spilled on the body, wash immediately with copious amount of water.

2. If water is spilled at the electrical points, inform the laboratory demonstrator

immediately. Do not try to do anything on your own.

3. If the pump does not run after the power is switched on, switch it off and inform

the laboratory demonstrator.

4. Handle glassware with care! In the event of breakage, clean the place immediately

and report to laboratory demonstrator. Do not touch the broken pieces with bare

hands.

5. Handle gas cylinders with care. If in doubt, ask the laboratory demonstrator. Donot open the main valve too rapidly. Ti may cause explosion.

6. Wear gloves when preparing and analyzing samples. Wear goggles at all times

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Results and Calculations

Packed Bed Data:

Height of column, h = 0.490m

Inner Diameter of packed bed column = 50 mmRaschig Rings

Inner Diameter = 6mm

Outer Diameter = 8mm

Length = 9mm

Other data

Molar volume of air at RTP = 24.0 L/mol

Molar concentration of water = 55.56 mol/L

Concentration of HCl used in titration = 0.1097 M

Concentration of NaOH used for back titration = 0.1022 M

Henrys Law constant, H A [6] = 29.82 atm L/mol

Diffusivity of CO 2 in solution, D A [6] = 1.92 x 10 -5 cm2 /s

Diffusivity of NaOH in solution, D B = 2.17 x 10 -5 cm2 /s

Experimental Results

RunInlet (Vol%)

Outlet (Vol%)

Sample 1 Sample 2 Average

CO 2 Air CO 2 Air CO 2 Air CO 2 Air

1 28.21 71.79 19.87 80.13 21.25 78.75 20.56 79.44

2 18.41 81.59 9.82 90.18 9.85 90.15 9.83 90.17

3 18.63 81.37 5.88 94.12 6.18 93.82 6.03 93.97

4 28.97 71.03 7.21 92.79 6.58 93.42 6.89 93.11

5 21.93 78.07 1.58 98.42 1.55 98.45 1.57 98.43

6 17.95 82.05 0.69 99.31 0.55 99.45 0.62 99.38Table 2: Composition of CO 2 and air in inlet and outlet gas flow measured using gaschromatography

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Run

Volume of HCl used for titration(Phenolphthalein)

Volume of HCl used for titration(Methyl Orange)

Sample 1 Sample 2 Average Sample 1 Sample 2 Average

1 3.30 3.20 3.25 1.80 1.80 1.802 3.80 3.70 3.75 3.40 3.30 3.35

3 6.20 6.10 6.15 3.10 3.30 3.20

4 8.20 8.50 8.35 6.35 6.10 6.23

5 9.80 9.70 9.75 4.15 4.10 4.13

6 11.90 11.90 11.90 3.80 3.40 3.60Table 3: Volume of HCl used for titration with the liquid sample drawn from absorption column.

Calculations

Run Inlet mole fractionof CO 2, yA,in Inlet mole ratio of

CO 2, Y A,in Outlet mole fraction

of CO 2, yA,out Outlet mole ratio

of CO 2, Y A,out

1 0.2821 0.3930 0.2056 0.2588

2 0.1841 0.2256 0.0983 0.1090

3 0.1863 0.2289 0.0603 0.0642

4 0.2897 0.4079 0.0689 0.0740

5 0.2193 0.2810 0.0157 0.0159

6 0.1795 0.2188 0.0062 0.0062Table 4: Tabulation of Mole Fraction / Ratio of CO 2 in inlet and outlet stream

Mole Fraction / Ratio of CO 2 in inlet and outlet stream

Sample Calculation

Using Sample 2,

Mole fraction of CO 2 (inlet), y A,in =100

41.18= 0.1841

Mole ratio of CO 2 (inlet), Y A,in =59.81

41.18= 0.2256

Mole fraction of CO 2 (outlet), y A,out =100

83.9= 0.0983

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Mole ratio of CO 2 (outlet), Y A,out =17.90

83.9= 0.1090

Calculation of NaOH concentration in feed tank

Volume of NaOH pipetted = 25 ml

Feed NaOH that was used in Week 1 was used for Runs 1 and 4 while feed NaOH thatwas used in Week 2 was used for Runs 2, 3, 5 and 6.

Week 1 Week 2

Reading 1 2 1 2

Final Reading (m l ) 37.10 37.00 27.50 27.50

Initial Reading (m l ) 0.00 0.00 0.00 0.00

Volume of HCL (m l ) 37.10 37.00 27.50 27.50

Average (m l ) 37.05 27.50

Concentration of NaOH in feed (mol/L)

0.1626 0.1205

Table 5: Titration results of volume of HCl used to neutratlise NaOH in the feeed tank

Sample Calculation

Using data from Week 1,

Number of moles of HCl required = 1097.0100005.37

= 310064.4 mol

Number of moles of NaOH pipetted = Number of moles of HCl required

= 310064.4 mol

Concentration of NaOH in the feed tank = 10002510064.4 3

= 1626.0 mol/L

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Table 6: Tabulation of Volumetric, Molar flow rate and Mole Fraction/Ratio of NaOH and deionised water ininlet stream

Calculation of mole ratio of NaOH from the inlet stream for different runs

Concentration of NaOH from feed tank = 0.1205 mol/L

Sample Calculation

Using data from Run 2:

Volumetric flow rate of H 2O = 0.55 L/min

Molar flow rate of H 2O = Molar volume of H 2O x Volumetric Flow rate (H 2O)

= 55.0181000

= 30.56 mol/min

Volumetric flow rate of NaOH = 0.05 L/min

Molar flow rate of NaOH = Concentration of NaOH from feed tank x Volumetric flow

rate of NaOH

= 05.01205.0

= 6.02 x 10 -3 mol/min

Mole fraction of NaOH, x B,in = Molar flow rate of NaOH / Total molar flow rate

=56.301002.6

1002.63

3

= 1.97 x 10 -4 mol NaOH/total mol

Run

Volumetricflow rate of

NaOH(L/min)

Volumetricflow rate of

H 2O(L/min)

Molar flowrate of NaOH

(mol/min)

Molar flowrate of H 2O(mol/min)

Mole fractionof NaOH

xB,in

Mole ratioof NaOH

XB,in

1 0.00 0.60 0.00 33.33 0.00 0.002 0.05 0.55 6.02 x 10 -3 30.56 1.97 x 10 -4 1.97 x 10 -4

3 0.10 0.50 1.20 x 10 -2 27.78 4.33 x 10 -4 4.34 x 10 -4

4 0.15 0.45 2.44 x 10 -2 25.00 9.75 x 10 -4 9.76 x 10 -4

5 0.20 0.40 2.41 x 10 -2 22.22 1.08 x 10 -3 1.08 x 10 -3

6 0.25 0.35 3.01 x 10 -2 19.44 1.55 x 10 -3 1.55 x 10 -3

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Mole ratio of NaOH, X B,in = 44

1097.11

1097.1

= 1.97 x 10 -4 mol NaOH/mol H 2O

RunMoles of HCl used

(Phenolphthalein)

Moles of HCLused

(Methyl Orange)

Moles of CO 2 (Outlet stream)

Moles of NaOHin sample

Moles of NaOH(Outlet Stream)

1 3.565 x 10 -4 1.975 x 10 -4 1.975 x 10 -4 1.591 x 10 -4 0

2 4.114 x 10 -4 3.675 x 10 -4 3.675 x 10 -4 4.388 x 10 -5 0

3 6.747 x 10 -4 3.510 x 10 -4 3.510 x 10 -4 3.236 x 10 -4 0

4 9.160 x 10 -4 6.829 x 10 -4 6.829 x 10 -4 2.331 x 10 -4 0

5 1.070 x 10 -3 4.525 x 10 -4 4.525 x 10 -4 6.171 x 10 -4 1.061 x 10 -4

6 1.305 x 10 -3 3.949 x 10 -4 3.949 x 10 -4 9.105 x 10 -4 3.995 x 10 -4 Table 7: Tabulation of Number of moles of NaOH and CO 2 in the outlet liquid stream

25 ml of the sample were drawn from the outlet stream for each run and 5 ml of NaOH

was added to the sample. The purpose of adding an additional amount of NaOH was to

ensure that all the CO 2 gas which was absorbed into the liquid stream reacted with NaOH

to form Na 2CO 3 instead of being dissolved in H 2O in the form of H 2CO 3. 2 rounds of

titration were carried with HCl using 2 indicators phenolphthalein followed by methyl

orange. As the pH range of phenolphthalein was between 8 to 10, the change in colour of phenolphthalein from pink to colourless at the end point indicated that the excess NaOH

was neutralized and also all the Na 2CO 3 reacted with HCl to form NaHCO 3. The volume

of HCl used neutralized NaOH and converted Na 2CO 3 to NaHCO 3. The chemical

reactions that occurred during the first round of titration are shown below.

A few drops of methyl orange was added and the mixture titrated with HCl until methyl

orange turned from yellow to orange. At the end point, HCl converted all the NaHCO 3 to

CO 2 and H 2O. The volume of HCl used reacted with all NaHCO 3. The difference in the

volume of HCl used in the 2 sets of titration allowed us to calculate the volume of NaOH

NaOH + HCl NaCl + H 2O

Na 2CO 3 + HCl NaHCO 3 + NaCl

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present in excess in the sample. The chemical reaction for the second set of titration is

shown below.

Calculation of mole ratio of NaOH and CO 2 from outlet stream for different runs

Sample Calculation

Using data from Run 2:

Moles of HCl used (1 st titration) = 1097.01000

75.3= 4.114 x 10 -4 mol

Moles of HCl used (2 nd titration) = 1097.01000

35.3= 3.675 x 10 -4 mol

Moles of CO 2 present in sample = 3.675 x 10 -4 molMoles of excess NaOH in sample = 4.114 x 10 -4 - 3.675 x 10 -4 = 4.388 x 10 -5 mol

Volume of NaOH added to sample = 5 ml

Moles of NaOH added to sample = 1022.01000

5= 5.11 x 10 -4 mol

Moles of NaOH in the sample obtained from outlet stream = 4.388 x 10 -5 - 5.11 x 10 -4

= -4.6719 x 10 -4

The negative value was obtained because some of the NaOH that was added to thesample reacted with the CO 2 that was dissolved in H 2O. Hence, the amount of NaOH in

the inlet liquid stream was not sufficient to result in total chemical absorption, thus

physical absorption of CO 2 into H 2O occurred in the lower part of the absorption column.

Since the number of moles of NaOH that was added to the sample was more than the

number of moles of excess NaOH in the sample, it can be deduced that there was no

NaOH in the sample collected from the outlet liquid stream (prior to the addition of

NaOH).

Mole ratio of CO 2 present in outlet stream, X A,out =

=

018

1000

1000

25

10675.3 4

Moles of CO 2(Moles of H 2O + Moles of NaOH)

NaHCO 3 + HCl NaCl + CO 2 + H 2O

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= 2.646 x 10 -4

Mole ratio of NaOH present in outlet stream, X B,out =

= 0

Table 8: Tabulation of mole ratio of NaOH and CO 2 in the outlet liquid stream

Calculation of overall mass transfer coefficient, K GAoa (without chemical reaction)

For Run 1, there was no NaOH introduced in the liquid stream. CO 2 that came into

contact with the liquid stream was absorbed by water (solvent) and hence the entire

column operates purely as a physical absorption zone. Hence, the overall mass transfer

coefficient, K GAoa can be calculated by the following equation.

L

mG

mX Y

mX Y

L

mG

LmG

hAP

GaK

ou t Bou t A

ou t Bin A

T

oGA

,,

,,1ln1

where

Cross sectional area of Column, A = 4

105023

= 310963.1 m2

Molar flow rate of carrier gas (air), G =24

0.1= 210167.4 mol/min

Molar flow rate of carrier liquid (water), L = 181000

60.0 = 33.33 mol/minTotal Pressure, P T = 1 atm

Height of column, h = 0.490 m

Mole ratio of NaOH in outlet liquid stream, X B,out = 0

Mole ratio of CO 2 in inlet gaseous stream, Y A,in = 0.3930

Mole ratio of CO 2 in outlet gaseous stream, YA, in = 0.2588

Run X A,in XA,out XB,in XB,out YA,in YA,out

1 0 1.422 x 10 -4 0 0 0.3930 0.2588

2 0 2.646 x 10 -4 1.971 x 10 -4 0 0.2256 0.1090

3 0 2.527 x 10 -4 4.336 x 10 -4 0 0.2289 0.0642

4 0 4.917 x 10 -4 9.754 x 10 -4 0 0.4079 0.0740

5 0 3.258 x 10 -4 1.084 x 10 -3 7.634 x 10 -5 0.2810 0.0159

6 0 2.843 x 10 -4 1.549 x 10 -3 2.876 x 10 -4 0.2188 0.0062

Moles of NaOH (Moles of H 2O + Moles of CO 2)

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Stripping Factor, L

mG=

L

GC H T A

=5556.0

10944.656.5582.29 4

= 2.071

Using the above found values, the overall mass transfer coefficient withoutchemical reaction was calculated and found to be:

K GAoa = 32.75 mol/ m 3.min.atm

Calculation of gas phase mass transfer coefficient, k gAa (with chemical reaction)

To obtain the gas phase mass transfer coefficient, k gAa, results from Run 6 were used. For

this run, as the concentration of NaOH fed into the column was very high, therefore therate of absorption will be controlled by the rate of mass transfer through the gas film, k gA .

The reaction can be assumed to be occurring only on the gas liquid interface and the

entire column assumed to be made up of the surface reaction zone.

I As Y A is quite significant, we cannot assume that the system is dilute. Hence, the

equation of the height of the surface reaction zone is given by the following equation:

)(ln ,,

,

,3 ou t Ain A

out A

in A

T gA

Y Y Y

Y

aAPk G

h

CB bulk film

PA

film bulk I

LiquidGas

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Cross sectional area of Column, A = 4

105023

= 310963.1 m2

Assuming ideal gas and room temperature conditions,

Molar flow rate of carrier gas (air), G =24

0.1 = 210167.4 mol/min

Total Pressure, P T = 1 atm

Height of column, h = 0.490 m

Mole ratio of CO 2 in inlet gaseous stream, Y A,in = 0.2188

Mole ratio of CO 2 in outlet gaseous stream, Y A,out = 0.0062

From the calculated values above, the gas phase mass transfer coefficient with chemical

reaction was calculated and found to be:k gAa = 163.28 mol/ m

3.min.atm

Calculation of liquid phase mass transfer coefficient without chemical reaction,k LA

0a

ak H

ak aK A L A

gAGA00

11

KGA oa = 32.75 mol/ m 3.min.atm

k gAa = 163.28 mol/ m 3.min.atm

Substituting the values above into the equation,

molatmm

Lm

mol Latm

ak aK

H ak

gAGA

A LA

.min.28.163

175.32

1

1000

1.82.29

11 3

3

0

0 = 1.221 min -1

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Calculation of the heights of the various reaction zones

a) Surface Reaction Zone

We first determined the zones inside the column by finding the critical mole ratio, X BC

which determines the end of the surface reaction zone and start of interior reaction zone.By comparing the liquid inlet and outlet mole ratio, X B,in and X B,out we will be able to

determine whether the surface reaction zone exists.

Mass transfer rate of A = (Mass transfer rate of B)/b

Since CO 2 in the gas phase is not dilute, T A

A A PY

Y P

1

b D

X DaC k

B

X aC D

b

X aC k

Y

Y aPk

A

BC BT LA BC T B BC T LB

A

AT gA0

3

3

1

3

3

0 1 A

A

T

T

B

A

LA

gA BC Y

Y b

C

P

D

D

ak

ak X

By material balance over the surface reaction zone,

BC out Bout A A X X bG L

Y Y ,,3

Solving for Y A3 and X BC by goal seek using Microsoft Excel for the various runs,

Run Y A, out XB, in XB,out XBc 1 0.2588 0 0 0.2022

2 0.1090 1.971 x 10 -4 0 8.529 x 10 -2

3 0.0642 4.336 x 10 -4 0 5.038 x 10 -2

4 0.0740 9.754 x 10 -4 0 5.837 x 10 -2

5 0.0159 1.084 x 10 -3 7.634 x 10 -5 1.302 x 10 -2

6 0.0062 1.549 x 10 -3 2.876 x 10 -4 5.713 x 10 -3

Table 9: Tabulation of Y A,out , X B,in , X B,out and X BC for the different runs

For Run 1, only physical absorption zone existed in the column as no NaOH was present

in the liquid stream.

For the remainder of the runs, as the mole ratio of NaOH in the inlet liquid stream, X B,in

was less than X BC , hence no surface reaction zone exist in the column.

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Since NaOH is present in the outlet liquid stream for Runs 5 and 6, hence the column is

made up of only the interior reaction zone in these 2 runs. For Runs 2 to 4, the column

constituted the interior reaction zone and the physical absorption zone. The table below

summarizes the reaction zones for each of the experimental runs.

Run Surface ReactionZone

Interior ReactionZone

Physical AbsorptionZone

1 X X

2 X

3 X

4 X

5 X X6 X X

Table 10: Tabulation of the different reaction zones present in the column during thedifferent experimental runs

Since the surface reaction zone is not present in all 6 runs, the interior reaction zone and

physical absorption zone was calculated for the different runs by the following equations:

b) Interior Reaction Zone

In the interior reaction zone, the mass transfer rate is given by the Hatta equation,

BT

A

A

B

A

AT

oGA

A AV X C b

H

D

D

Y

Y PaK

dt

dN

V r

1

1----- (1)

By material balance around the interior reaction zone,

)( ,, out A Ain B B Y Y LbG

X X ----- (2)

By substituting (2) into (1) and then performing integration with the use of MATLAB toobtain the height for interior reaction zone (See Appendix A),

2

,

2

A

ou t A

Y

Y AV

A

r

dY Gh

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c) Physical Absorption Zone

The height of the physical absorption zone was calculated using following equation:

L

mG

mX Y

mX Y

L

mG

L

mGaAPK

Gh

B A

Bin A

T o

GA 22

2,3 1ln

1

The table below summaries the heights of the reaction zones for the 6 runs.

Run Surface ReactionZone, h 1 (m)Interior Reaction

Zone, h 2 (m)Physical Absorption

Zone, h 3 (m)Total Height

(m)

1 0 0 0.4900 0.4900

2 0 0.0238 0.1465 0.1703

3 0 0.0787 -0.0234 0.0787

4 0 0.3526 -0.0740 0.3526

5 0 0.4068 0 0.4068

6 0 0.8037 0 0.8037

Table 11: Tabulation of the heights of the different reaction zones in the column for thedifferent experimental runs

Calculation of Enhancement factor, E

For instantaneous reaction, E is given by:

ave A A

Aave B B

Pb D

H C D E

,

,1

Since C B is not constant along the packed column, the enhancement factor changes along

the column. Therefore, average C B is used to calculate the enhancement factor for each

run.

Sample Calculation

Using date from Run 2,in BC , = in BT X C ,

= 55.56 x 1.971 x 10 -4

= 1.095 x 10 -2 mol/L

ou t BC , = out BT X C ,

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= 55.56 x 0

= 0 mol/L

ave BC , = 2,, out Bin B C C

= 5.475 x 10 -3 mol/L

in AP , =in A

in AT Y

Y P

,

,

1

=2256.01

2256.01

= 0.451 atm

out AP , =out A

out AT

Y

Y P

,

,

1

=1090.01

1090.01

= 0.218 atm

ave AP , = 2,, ou t Ain A PP

= 0.335 atm

Substituting the values above into the equation,

E = 1.276

Run C B,in C B,out C B,ave P A,in P A,out P A,ave E

1 0 0 0 0.786 0.518 0.652 1.000

2 1.095 x 10 -2 0 5.475 x 10 -3 0.451 0.218 0.335 1.276

3 2.409 x 10 -2 0 1.204 x 10 -2 0.458 0.128 0.293 1.693

4 5.419 x 10 -2 0 2.710 x 10 -2 0.816 0.148 0.482 1.948

5 6.022 x 10 -2 4.241 x 10 -3 3.223 x 10 -2 0.562 0.032 0.297 2.830

6 8.603 x 10 -2 1.598 x 10 -2 5.100 x 10 -2 0.438 0.012 0.225 4.819

Table 12: Tabulation of the Enhancement Factor for the different runs

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Comparing the overall improvement in absorption due to chemical reaction interms of the effectiveness factor

The percentage of CO 2 absorbed as the flow rate of NaOH was increased in the inlet

liquid stream was used as an indicator for the improvement in absorption rate. The

equation is given below:

Percentage of CO 2 absorbed = 100,

,,

in A

out Ain A

Y

Y Y

The table and graph below summarise the relationship between the percentage of CO 2

absorbed and the effectiveness factor between each run. It was observed that the

percentage of CO 2 absorbed increased as the effectiveness factor increased. However, the

percentage of CO 2 absorbed reached a saturation point as more NaOH was introduced.

Run Enhancement Factor Percentage of CO 2 absorbed (%)

1 1.000 34.14

2 1.276 51.67

3 1.693 71.96

4 1.948 81.85

5 2.830 94.33

6 4.819 97.15Table 13: Extent of CO 2 absorption with respect to Enhancement Factor

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Graph of Percentage of CO 2 absorbed against Enhancement Factor

0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

0.000 1.000 2.000 3.000 4.000 5.000 6.000

Enhancement Factor

P e r c e n

t a g e o

f C O

2 a

b s o r b e

d

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Discussion

Based on the data of the experiment collected, the inlet CO 2 concentration is much higher

than 5 volome% .

Run Inlet mole fraction of CO 2, yA,in Inlet mole ratio of CO 2, Y A,in

1 0.2821 0.3930

2 0.1841 0.2256

3 0.1863 0.2289

4 0.2897 0.4079

5 0.2193 0.2810

6 0.1795 0.2188

Since the system can be assumed to be dilute only when the concentration of the

component is less than 5% mole fraction, the system in this experiment was considered as

non-dilute. Therefore, the calculations performed above were based on mole ratio rather

than mole fraction.

Height of each zone

1. The accuracy of the height calculated for the three zones in each runFrom the calculated result shown in the above table, there are negative numbers in run 3

and run 4. Based on the analysis of the equation used for computing the physical

absorption zone, the negative column height implied that the stripping took place instead

of absorption. This is against the law of thermodynamics and is thus flawed. However,

given that the values of the heights are close to zero, the values obtained could be a result

of experimental error during the titration or collection of sample. Furthermore, from

Table 9, it can be seen that there was no NaOH in the liquid outlet for runs 3 and 4. As

we do not have information of the concentration of NaOH at different points along the

column, it could be true that the end of the interior reaction zone could be near the end of

the entire column. Hence, the height of the physical absorption zone could be

insignificant or close to zero.

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The calculations performed above were based on the assumption that Run 6 had only

surface reaction zone present as Run 6 had the highest inlet NaOH concentration, 1.55

10 -3 % mole fraction. With this assumption, k gAa, the mass transfer coefficient in gas

phase with chemical reaction can be calculated. Thus the following calculations can be

done to calculate the heights of each zone. However, from the results calculated, we

found out that in run 6, there is actually no surface reaction zone present because the inlet

NaOH concentration is less than X BC , the critical concentration of NaOH for surface

reaction zone, 5.713 x 10 -3. There is only interior surface reaction zone present in Run 6.

This shows that our assumption was actually wrong. As a result, the calculated k gAa was

incorrect and the subsequent calculation done based on that k gAa value was also

inaccurate.

Another possible source of inaccuracy in the calculation was the fluctuating flow rate.

Even though we fixed the flow rate of CO 2 and air to be 0.2 L/min and 1 L/min, the inlet

concentration of CO 2 measured from gas chromatography was not constant. For both

water and NaOH, the rotameter reading did not stay at the level we adjusted. It was

difficult to attain a truly steady state. Therefore, the mass transfer coefficient k gAa, k LAa,

and thus K gAa were actually not constant for each run. Yet, we used the overakl mass

transfer coefficient calculated from run 6 result for the calculations in the other runs. Thisresulted in some inaccuracy in our final results.

From theory, we know that by setting CO2 flow rate to 0.2 L/min, and air flow rate to 1.0

L/min, the volume% obtained from gas chromatography should read around 16.7% for

CO 2. However, from the data collected, the inlet volume percent of C0 2 was about 30%,

which is close to double of what was expected. One reason could be due to the

excessively high pressure from the CO 2 gas tank, which resulted in a highly concentrated

CO 2 flow stream into the inlet of the column. Thus, the flowrate of CO 2 can be lowered

in the experiment. Another reason could be the poor mixing of the CO 2 and air streams as

the samples were taken immediately after they have mixed. This could lead to inaccurate

mass transfer calculations.

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Another possible explaination for the deviation of the column height could be due to poor

initial liquid distribution and the irrigation quality from the distributor, resulting in the

deterioration of column efficiency and the mass transfer performance of the packed

column. A less efficient mass transfer would result in a longer column height to achieve a

desired separation, however since calculation was done based on the outlet mole ratio of

the liquid and gases, This could justify the smaller column heights calculated and hence a

shorter column.

One more assumption we made was that the mass transfer in the gas absorption is based

on film theory and that the interfacial film area is stagnant. However this assumption is in

doubt. Many researches and investigations have modeled the mass transfer process using

different theories in which a non-stagnant interfacial film area is involved.

There were also other experimental errors involved in the titration process such as the

parallax errors in the reading of the liquid level during titration. This will be discussed

more in error analysis section.

2. Relationship of height of each zone and the NaOH concentration

As NaOH flow rate was increased, it was observed that the interior reaction zone

increased and the physical absorption zone decreased. This goes well with the theory.

With more NaOH present, there will be higher C B concentration at the interface. The

RunSurface Reaction

Zone, h 1 (m)

Interior Reaction

Zone, h 2 (m)

Physical Absorption

Zone, h 3 (m)

Total Height

(m)

1 0 0 0.4900 0.4900

2 0 0.0238 0.1465 0.1703

3 0 0.0787 -0.0234 0.787

4 0 0.3526 -0.0740 0.3526

5 0 0.4068 0 0.4068

6 0 0.8037 0 0.8037

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reaction plane will shift towards the interface. The surface reaction zone if present should

increase with increasing NaOH flowrate based on theory.

Mass transfer coefficients

KGAoa 32.7461231

kgAa 163.2764288kLA

oa 1.221461174

Comparing the mass transfer coefficients shown above, KLAoa is much smaller than kgAa .

From the calculations, it can be observed that the resistance to mass transfer in the liquid

film is much more significant than the resistance to mass transfer in the gas film. Hence

the mass transfer of CO 2 is liquid phase controlled. Based on the experiment conducted

by Brettschneider et al, t he fraction of liquid mass transfer resistance for CO2

varies from

40 up to 90%. The fraction of liquid mass transfer resistance for CO 2 mass transfer

resistance is located almost completely on the liquid side. However, its contribution was

reduced to less than 40% in the more reactive sections of the column. The experimental

results is illustrated in the diagram below. From the diagram, it can be observed that the

resistance to mass transfer on the liquid phase actually varies at different sections of the

column. Furthermore, the high percentage of mass transfer resistance occurring in the

liquid side further validates our calculation for KLAoa.

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Compare the overall improvement in absorption due to chemical reaction in termsof effective factor as a function of the operating conditions .

Graph of Percentage of CO 2 absorbed against Enhancement Factor

0.00

10.00

20.00

30.00

40.00

50.00

60.00

70.00

80.00

90.00

100.00

0.000 1.000 2.000 3.000 4.000 5.000 6.000

Enhancement Factor

P e r c e n t a g e o

f C O

2 a

b s o r b e

d

Assumptions made in the experiment

1. The rate of reaction in the liquid is instantaneous relative to the rate of mass

transfer. Thus, the overall rate will only be determined by mass transfer rate and

independent of the rate of reaction.

The reaction between sodium hydroxide and carbonic acid is an acid base reaction.

The neutralization reaction involves the combination of a H + and an OH - ion.

Reaction between ions requires very low activation energy and thus the rate of reaction is relatively fast. In an instantaneous reaction, the reactants cannot

coexist together can thus the reaction occurs in a plane.

For an instantaneous reaction, reaction occurs in a plane anywhere within the

liquid film and its location depends on the relative mass transfer rates of the

absorbed carbon dioxide and NaOH.

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2. Henrys law is applicable.

This assumption in turn assumes dilute concentration and partial pressure of

carbon dioxide in the gas and aqueous phase. Since the modified Henrys law

constant is also a function of temperature and pressure, temperature and pressure

is assumed to be constant throughout the column.

3. The chemical reaction takes place only in the liquid phase, no reaction occurs in

the gaseous phase.

Since NaOH is a non-volatile solute, we can safely assume that none of it escapes

into the gaseous phase. And since carbon dioxide is only weakly soluble in water,

there will be no carbonic acid in the moisture of the gas phase. NaOH can only

react with dissolved carbon dioxide (carbonic acid).

4. There is no evaporation of water into the gas stream. With this assumption, weassume the inert liquid flow rate L to be constant. Similarly, we assume that other

gases in the gas stream such as oxygen do not dissolve in the liquid stream so that

inert gas flow rate G is constant.

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Error Analysis

1. Steady state was assumed to be reached as long as two consistent readings for the

gas samples were obtained. In reality, however, both the gas and liquid flow rates

were fluctuating from time to time, and not constant. Therefore, steady state wasnot really attained, leading to erroneous data recorded.

2. It was found out that that the assumption that Run 6 had only surface reaction

zone was incorrect as the concentration of NaOH entering was too low for that to

take place. It was based on the assumption that Kga was calculated, leading to a

rather inaccurate value of Kga obtained. That value made the subsequent

calculations incorrect.

3. Our system was not operated based on dilute conditions. Therefore, using Henrys

constant in the calculations was incorrect. For a non-dilute system, the

equilibrium curve is not a straight line and therefore Henrys constant does not

really apply.

4. The inlet gas ratios for all the runs showed inconsistent readings. This shows the

possibilities of contamination of gas syringes and atmospheric air taken in

together with the gas samples when drawing samples. Therefore, the values

obtained from the gas chromatograph did not reflect the actual composition of the

gas samples.

5. Possible errors on titration include parallax errors when taking readings from the

burettes and human judgment errors when observing colour change. Improper

washing of the apparatus could lead to contamination of reagents and liquid

samples. As a result, inaccurate data may have been be taken.

6. NaOH solution on the feed tank was not mixed thoroughly, therefore the sample

taken out from the feed tank for titration may not have at the same concentration

as that fed into the column.7. The mass transfer operation may not be explained entirely only by the two-film

theory like what was assumed. Other possible theories that can complement the

two-film theory include penetration theory and surface rejuvenation theory.

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8. As the inlet CO 2 concentrations were very high in the experiment, the assumption

that CO 2 does not dissolve in the liquid solvent might not hold. CO 2 concentration,

when high enough, can dissolve into the liquid solvent to a certain extent.

9. Due to the enthalpy of chemical reaction and different environmental conditions

between week one and two of experiment, temperature and pressure may not be

constant unlike what we had assumed.

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Conclusion

From this experiment, we can conclude that the presence of the three absorption zones

(namely surface reaction zone, interior reaction zone and physical absorption zone) are

directly related to the concentration of NaOH entering the packed tower since theflowrates of CO 2 and air are not changed. In run 1 (pure water used only), the column is

solely operating under physical absorption zone. The mass transfer of the liquid phase is

controlling in this whole experiment.

When a low concentration of NaOH is used (run 2 to 5), the interior reaction zone as well

as the physical absorption zone are present. As the NaOH concentration is increased, the

reaction with carbonic acid formed moves the reaction nearer to the gas-liquid surface.

k LAa increases with the concentration of NaOH in the feed. k gAa is taken to be constant

and therefore the overall gas mass transfer coefficient increases with concentration of

NaOH. The mass transfer resistance in the liquid phase decreases with increase in

concentration of NaOH.

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References

1. O. Brettschneider et al. Separation and Purification Technology 39 (2004) 139 159) Experimental investigation and simulation of the chemical absorption in a

packed column for the system NH 3 CO 2 H2S NaOH H2O

2. Cooper, A.R. and G.V. Jeffreys, Chemical Kinetics and Reactor Design ,Prentice-Hall, Inc., 1971.

3. Geankoplis, C.J., Transport Processes and Unit Operations , 3 rd Ed., Prentice-Hall, Inc., 1993.

4. Levenspiel, O., Chemical Reaction Engineering , 3 rd Ed., Wiley, Inc. 1999.

5. Treybal, R.E., Mass-Transfer Operations , 3 rd Ed., McGraw-Hill, 1981.

6. Perry, R.H. and Green, D.W., Perrys Chemical Engineers Handbook , 7 th Ed.,McGraw-Hill, 1997.

7. Danckwerts, P.V. and Sharma, M.M., I.CHEM.E.REVIEW SERIES , TheChemical Engineer. 1996.

8. Beavon, R., Determination of Sodium Hydroxide and Sodium Carbonate in the Presence of One Another , Retrieved 05 September, 2006 fromhttp://www.rod.beavon.clara.net/mixedind.htm

9. Astarita, G., Mass Transfer with Chemical Reaction , Elsevier PublishingCompany Limited, 1967

10. Hikita, H., Asai,S. and Takatsuka,T., "Absorption of carbon dioxide into aqueoushydroxide and sodium carbonate- bicarbonate solutions", The ChemicalEngineering Journal, 11 (1976) 131-141

11. http://www.nzifst.org.nz/unitoperations/conteqseparation8.htm

12. Chemical Engineering Science 55(2000)

http://www.rod.beavon.clara.net/mixedind.htmhttp://www.rod.beavon.clara.net/mixedind.htmhttp://www.nzifst.org.nz/unitoperations/conteqseparation8.htmhttp://www.nzifst.org.nz/unitoperations/conteqseparation8.htmhttp://www.nzifst.org.nz/unitoperations/conteqseparation8.htmhttp://www.nzifst.org.nz/unitoperations/conteqseparation8.htmhttp://www.rod.beavon.clara.net/mixedind.htm

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Appendix A

M-File for calculating the height of the Interior reaction zone and physical absorption zone

%Define Variables% P_T = 1; K_GAa = 32.7461231; Db = 2.170E-05; Da = 1.920E-05; HA = 29.82; b = 2; CT = 1000/18; Xb_in = [0.000E+00 1.971E-04 4.336E-04 9.754E-04 1.084E-03 1.549E-03]'; Ya_out = [0.2588 0.1090 0.0642 0.0740 0.0159 0.0062]'; G = 4.167E-02; L = 33.33333333;

%Solve for Ya_2 by material balance% Ya_2 = Ya_out + (L/(b*G))*Xb_in;

syms Ya

Xb = Xb_in - (b*G/L)*(Ya - Ya_out); %Integrating function with respect to Ya to obtain height of theinterior% %reaction zones for each run%

fprintf( 'Heights of interior reaction zones for Runs 1 to 6:\n' ); f1 = K_GAa*(P_T*(Ya/(1+Ya) + (Db/Da)*(HA/b)*CT*Xb(1))); height_1 = G*int(f1,Ya,Ya_out(1),Ya_2(1)); height1 = eval(height_1); f2 = K_GAa*(P_T*(Ya/(1+Ya) + (Db/Da)*(HA/b)*CT*Xb(2))); height_2 = G*int(f2,Ya,Ya_out(2),Ya_2(2)); height2 = eval(height_2);

f3 = K_GAa*(P_T*(Ya/(1+Ya) + (Db/Da)*(HA/b)*CT*Xb(3))); height_3 = G*int(f3,Ya,Ya_out(3),Ya_2(3)); height3 = eval(height_3);

f4 = K_GAa*(P_T*(Ya/(1+Ya) + (Db/Da)*(HA/b)*CT*Xb(4))); height_4 = G*int(f4,Ya,Ya_out(4),Ya_2(4)); height4 = eval(height_4);

f5 = K_GAa*(P_T*(Ya/(1+Ya) + (Db/Da)*(HA/b)*CT*Xb(5))); height_5 = G*int(f5,Ya,Ya_out(5),Ya_2(5)); height5 = eval(height_5);

f6 = K_GAa*(P_T*(Ya/(1+Ya) + (Db/Da)*(HA/b)*CT*Xb(6))); height_6 = G*int(f6,Ya,Ya_out(6),Ya_2(6)); height6 = eval(height_6);

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h_interior = [0 height2 height3 height4 height5 height6]'

%Calculating physical absorption zone% m = CT*HA/P_T; A = 1.963E-03;

Ya_in = [0.3930 0.2256 0.2289 0.4079 0.2810 0.2188]'; fprintf( 'Heights of physical absorption zone for Runs 1 to 6:\n' ); for i = 1:1:4

h(i) = (G/(K_GAa*A*P_T*(1-m*G/L)))*log((1-m*G/L)*(Ya_in(i)/Ya_2(i))+ m*G/L);

h(5) = 0; h(6) = 0;

end h = h'

Matlab Program

Heights of interior reaction zones for Runs 1 to 6:

h_interior =

00.02380.07870.35260.40680.8037

Heights of physical absorption zone for Runs 1 to 6:

h =

0.49060.1468

-0.0233-0.0738

00