gas absorption report.pdf

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UEMK2411 CHEMICAL ENGINEERING LABORATORY I GROUP 09

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TITLE OF EXPERIMENT

Gas Absorption

OBJECTIVE(S)

The objective of this experiment is to determine the mass transfer coefficient

of oxygen through wetted wall absorption column.

INTRODUCTION

Gas absorption (also known as scrubbing) is an operation in which a gas

mixture is contacted with a liquid for the purpose of preferentially dissolving one or

more components of the gas mixture and to provide a solution of them in the liquid.Therefore we can see that there is a mass transfer of the component of the gas from

the gas phase to the liquid phase. The solute so transferred is said to be absorbed by

the liquid. In gas desorption (or stripping), the mass transfer is in the opposite

direction, i.e. from the liquid phase to the gas phase.

The principles for both systems are the same. The process of gas absorption

involves the diffusion of solute from the gas phase through a stagnant or non-

diffusing liquid. Absorption can be either physical or chemical. In physical

absorption, the gas is removed because it has greater solubility in the solvent than

other gases. In chemical absorption, the gas to be removed reacts with the solvent

and remains in solution. For irreversible reactions, the resulting liquid must be

disposed of, whereas in reversible reactions, the solvent can be regenerated. Thus,

reversible reactions are often preferred. Chemical absorption usually has a much

more favourable equilibrium relationship than physical absorption (solubility of most

gases is usually very low) and is, therefore, preferred. Both absorption and stripping

can be operated as equilibrium stage operations with contact of liquid and vapour. In

both absorption and stripping a separate phase is added as the separating agent.

MATERIALS AND EQUIPMENT

Deoxygenated water Nitrogen gas, N 2 Oxygen gas, O 2


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Wetted-wall absorption column Stop watch

RESULTS AND CALCULATIONS

Information given:

Column diameter

Column height

Wetted perimeter Gas- liquid interface area

Oxygen diffusivity

Theoretically,

( )

[ ]

An Excel spread sheet is attached with the complete results table.


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Results Tabulation For Gas Absorption

Time Interval 5 min

Concentration mg/L

Gas Water 1st 2nd 3rd Average 1st 2nd 3rd Average Gas liq4 0.25 0.21 0.19 0.22 5.61 5.30 5.33 5.41 6 0.20 0.08 0.07 0.12 5.43 5.43 5.59 5.488 0.08 0.07 0.05 0.07 5.80 5.66 5.71 5.72

10 0.05 0.05 0.05 0.05 5.96 5.97 5.99 5.974 0.05 0.04 0.13 0.07 5.80 5.70 5.62 5.716 0.06 0.04 0.04 0.05 5.53 5.61 5.50 5.558 0.04 0.04 0.04 0.04 5.59 5.33 5.57 5.50

10 0.04 0.05 0.04 0.04 5.75 5.79 5.81 5.784 0.04 0.05 0.05 0.05 5.44 5.50 5.17 5.37

6 0.05 0.05 0.07 0.06 5.20 5.42 5.36 5.338 0.06 0.05 0.04 0.05 5.40 5.30 5.46 5.3910 0.04 0.04 0.04 0.04 5.75 5.87 5.84 5.82

Temperature deg C

Gas Water 1st 2nd 3rd Average 1st 2nd 3rd Average4 31.80 32.60 30.60 31.67 30.10 30.60 31.20 30.636 34.20 34.60 34.80 34.53 31.60 32.10 32.40 32.038 35.00 34.70 34.60 34.77 32.70 32.90 32.90 32.83

10 34.40 34.30 34.10 34.27 32.90 32.90 32.80 32.874 33.90 33.70 34.60 34.07 32.40 32.10 32.00 32.176 34.70 35.30 35.50 35.17 32.20 32.90 33.10 32.738 36.90 35.90 35.80 36.20 33.40 33.80 33.70 33.63

10 35.60 35.30 35.20 35.37 33.80 33.80 33.70 33.774 34.90 34.60 34.80 34.77 33.40 32.90 32.80 33.036 35.60 36.00 35.80 35.80 32.90 33.50 33.60 33.338 36.00 36.00 35.90 35.97 33.70 34.00 33.90 33.87

10 35.70 35.50 35.40 35.53 33.90 33.80 33.70 33.80

Flowrate (L/h)

Flowrate (L/h) AI1 AI2

AI2AI1

100

80

60

100

60

80


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T average viscosity density wetting rate Re Gas Water in out deg C cm^2/s g/cm^3 cm^2/s C_i C_o

4 7.29 7.40 31.15 0.0077 0.9949 0.1040 13.36 7.08 1.996 7.03 7.25 33.28 0.0074 0.9942 0.1561 21.05 6.91 1.778 7.01 7.18 33.80 0.0073 0.9941 0.2081 28.39 6.95 1.45

10 7.05 7.17 33.57 0.0073 0.9941 0.2601 35.30 7.00 1.204 7.07 7.24 33.12 0.0074 0.9943 0.1040 13.98 7.00 1.546 6.98 7.19 33.95 0.0073 0.9940 0.1561 21.37 6.94 1.648 6.91 7.11 34.92 0.0071 0.9937 0.2081 29.11 6.87 1.61

10 6.97 7.09 34.57 0.0072 0.9938 0.2601 36.10 6.92 1.314 7.01 7.16 33.90 0.0073 0.9940 0.1040 14.23 6.97 1.796 6.94 7.13 34.57 0.0072 0.9938 0.1561 21.66 6.88 1.818 6.92 7.09 34.92 0.0071 0.9937 0.2081 29.11 6.87 1.70

10 6.95 7.09 34.67 0.0071 0.9938 0.2601 36.18 6.91 1.27

N_A K_f K_L Sh Re ShGas Water g/s.m^2 g.s/cm^2 cm/s Gas Water4 6.0097E 06 1.4996E 06 8.5378E 07 3.07 4 13.36 3.076 9.3094E 06 2.4650E 06 2.4793E 06 8.93 6 21.05 8.938 1.3083E 05 3.7250E 06 3.7472E 06 13.49 8 28.39 13.49

10 1.7125E 05 5.2037E 06 5.2343E 06 18.84 10 35.30 18.844 6.5146E 06 1.8095E 06 1.8199E 06 6.55 4 13.98 6.556 9.5407E 06 2.5976E 06 2.6132E 06 9.41 6 21.37 9.418 1.2621E 05 3.4823E 06 3.5043E 06 12.62 8 29.11 12.62

10 1.6595E 05 4.9203E 06 4.9508E 06 17.82 10 36.10 17.824 6.1561E 06 1.6164E 06 1.6260E 06 5.85 4 14.23 5.856 9.1417E 06 2.4101E 06 2.4250E 06 8.73 6 21.66 8.738 1.2343E 05 3.3334E 06 3.3545E 06 12.08 8 29.11 12.08

10 1.6711E 05 5.0144E 06 5.0457E 06 18.16 10 36.18 18.16

Flowrate (L/h)

Flowrate (L/h) Satura ted Concentration Difference in concentration

60

80

100

Flowrate (L/h)

60

80

100

60

80

100


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y = 0.7088x 6.3003 R = 0.9983

0.00

5.00

10.00

15.00

20.00

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00

S h

Re

Sh VS Re for Air Flow 60

Air 60 Linear (Air 60)

y = 0.49R

0.00

5.00

10.00

15.00

20.00

0.00 5.00 10.00 15.00 20.00 25.00

S h

Re

Sh VS Re for Air Flow


y = 0.5481x 2.6584 R = 0.9617

0.00

5.00

10.00

15.00

20.00

0.00 5.00 10.00 15.00 20.00 25.00 30.00 35.00 40.00

S h

Re

Sh VS Re for Air Flow 100



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y = 1.8573x 3.6142 R = 0.983

0.0000

0.5000

1.0000

1.5000

2.0000

2.5000

3.0000

3.5000

0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

l n S h

ln Re

ln Sh VS ln Re for Air Flow 60


0.0000

0.5000

1.0000

1.5000

2.0000

2.5000

3.0000

3.5000

0.0000 0.5000 1.0000 1.5000 2.0000

l n S h

ln Re

ln Sh VS ln Re for Air

Air 80 Linear (A

y = 1.1766x 1.4017 R = 0.9768

0.0000

0.5000

1.0000

1.5000

2.0000

2.5000

3.0000

3.5000

0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

l n S h

ln Re




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Sample calculation for the set of air flow rate, and water flow rate, . At this set of flow rate, we obtained the following results:

First, we get saturation concentration by using the following formula:

Then, we calculate the average temperature, :

After that, is used to find the viscosity, and also the density of liquid, .

Viscosity of water,

Density of water,


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After getting the viscosity and the density of water, now we are going to find wettingrate, .

( )

After getting the wetting rate, we can get the Reynolds number, for the water.

Next, we are going to find the rate of mass transfer,

Now, once we have , we can calculate for

After getting , we are able to get Sherw oods number, .


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The next step is to both of the and number and a graph of was plotted.

DISCUSSION

The diagram below shows the schematic diagram for this experiment.

Based on our experiment, all the results are tabulated in the tables and graphs

are plotted. From the results that we collected, we found that when the water flow

rate increases at the constant air flow rate subsequently it will reduce the inlet of

oxygen concentration. The relationship between the water flow rate against the

wetting rate and the Reynolds number can be conclude that when the water flow rate

increases, both the wetting rate and the Reynolds number will also increases as well.Furthermore, the increment of the mass transfer coefficient, k L will affect the rate of


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mass transfer. The relationship between the Sherwood number and Reynolds number

are shown in the calculation part and graphs of it are plotted. From the calculation,

we know that the when water flow rate increases, the Sherwood number (Sh) also

increases.

Graphs of LSh )ln( versus Lln(Re) for different air flow rate are plotted below.

Graph of was plotted. The plotting of the graphs is to determine theslope of the graph. From the graphs, it is showed that the when the water flow rate

increases, it lead the graph of increases linearly. Graphs of differentflow rate showed the same effect when the water flow rate increases in this

experiment.

y = 1.8573x - 3.6142R = 0.983

0.0000

0.5000

1.0000

1.5000

2.0000

2.5000

3.0000

3.5000

0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

l n S

h

ln Re




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Air flow rate = 60 l/h Air flow rate = 80 l/h Air flow rate = 100 l/h

Re value 0.9991 0.9997 0.9999

Table 2: Re2 for different air flow rate

From the plotted graphs, it showed that the Re 2 value increases when the air flow rate

increases.

In this experiment, the water diffusion is the mass transfer. From all the

information given in the lab manual and from the reading of the machine, we only

y = 1.0239x - 0.8555R = 0.9808

0.0000

0.5000

1.0000

1.5000

2.00002.5000

3.0000

3.5000

0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

l n S

h

ln Re



y = 1.1766x - 1.4017R = 0.9768

0.0000

0.5000

1.0000

1.5000

2.0000

2.5000

3.0000

3.5000

0.0000 0.5000 1.0000 1.5000 2.0000 2.5000 3.0000 3.5000 4.0000

l n S

h

ln Re




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know the concentration of the dissolved oxygen and the temperature. All this value

leads us to calculate the mass transfer coefficient when the driving force is expressed

in the term of concentration. We know that gas absorption essentially involved the

transfer of materials from the gas phase to the liquid phase. It is also defined as the

operation in which a gas mixture is contacted with a liquid for the purpose of

preferentially dissolving one or more components of the gas mixture and to provide a

solution of them in the liquid. The gaseous component is said to be absorbed by the

liquid. On this gas absorption, mass transfer is occurring where there is a net

movement of mass from one location to another because of a variance in absorption.

In fact, mass transfer is strongly influenced by molecular spacing, diffusion occurs

more readily in gases than in liquid and more readily in liquids than in solid

The accuracy of this experiment is affected by a few errors. The diffusion

coefficient of oxygen is assumed to be constant throughout the experiment. In reality,

the diffusion coefficient depends on both pressure and temperature. In adjusting the

flow rate of air and water, parallax error might occur when readings are taken. Thus,

the results are affected. To reduce this human error, few readings should be taken at

the eye level and only consider the average reading when calculation is to be done.

During the experiment, there is no precise value to determine the amount of nitrogengas needed to produce a steady stream of nitrogen bubbles to be fed into

deoxygenated column. Therefore, when the valve is being turned on and off, the flow

of nitrogen gas will be different and eventually affect the subsequent results above.

CONCLUSION

As a conclusion, the objective was achieved. The Re 2 value increases when

the air flow rate increases. From the results of this experiment, Reynolds number is

proportional to the Sherwood number. Thus, we can conclude that the mass transfer

coefficient is also proportional to the water flow rate.


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REFERENCES

1. ChE 382: Unit Operations Laboratory. (n.d.). Retrieved July 3, 2011, from

ChE 382: Unit Operations Laboratory:

http://www.uic.edu/depts/chme/UnitOps/CO2.pdf

2. (2011). Mass Transfer. In E. Henley, J. Seader, & D. Roper, Separation

Process Principles (pp. 91-149). Asia: John Wiley & Sons Pte Ltd.

3. Incropera, F. P., Dewitt, D. P., Bergman, T. L., & Lavine, A. S. (2005). Mass

Transfer by Diffusion. In F. Incropera, D. Dewitt, T. Bergman, & A. Lavine,

Fundamentals of Heat and Mass Transfer (pp. 879 - 916). Asia: John Wiley

& Sons Inc.

4. MASS TRANSFER IN GAS ABSORPTION & DIFFUSION . (n.d.). Retrieved

July 3, 2011, from http://www.separationprocesses.com:

http://www.separationprocesses.com/Absorption/GA_Chp02.htm

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