Contents

Introduction.....................................................................................................................................2
Experimental Procedure...................................................................................................................2
Results.............................................................................................................................................3
Discussion........................................................................................................................................6
Conclusion.......................................................................................................................................6
Appendix.........................................................................................................................................8

List of Tables
Table 1: Experimental Data at Different Air and Water Flow Rates..............................................3
Table 2: Calculated Data at Different Air and Water Flow Rates...................................................4
Table 3: The Values of n and R2 of Different Air Flow Rates........................................................6
Table 4: Data for Sample Calculation.............................................................................................7
Table 5: Interpolation Csat ,∈¿ at Tin=29 .10 ℃...........................................................................7
Table 6: Interpolation Psat ,∈¿ at Tin=29 .10 ℃...........................................................................8
Table 7: Interpolation kinematic viscosity , ν At Tavg=28 ℃.........................................................8
Table 8: Calculated Data at 4.0 L/h of Water Flow Rate................................................................9
Table 9: Interpolation Density , ρ at Tavg=29 . 55℃......................................................................9

List of Graphs
Graph 1: Air Flow rate at 60 L/h.....................................................................................................5
Graph 2: Air FLow Rate at 120 L/h.................................................................................................5
Graph 3: Air Flow Rate at 180 L/h..................................................................................................5
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Introduction
Gas absorption is defined as a widely-used industrial operation where a gas mixture is mixed
with a suitable liquid, thus, enabling transfer of a component between the gas and liquid stream.

Through gas absorption, the gas mixture can easily be separated and the components in gas
mixture that are unnecessary or harmful can be effectively removed. Furthermore, some useful
components can be recovered depending on the gaseous mixture and liquid used for the
absorption.

The equipment used for our experiment is a wetted wall column and it is designed to conduct
studies on oxygen absorption from air into deoxygenated water, which is prepared using nitrogen
gas. Thus, the objectives for this experiment are:

1. To study the solubility of oxygen in pure water at different temperatures.

2. To determine the mass transfer coefficients from the experimental data.

Experimental Procedure
Start up

1. Main switch is turned on.

2. Valve VI is connected to the nitrogen gas cylinder and set to a pressure of 0.5 – 1 bar.
3. Tank (D1) is filled with distilled water.
4. Pump G1 is switched on and valve V3 is adjusted carefully and slowly to fill the
deoxygenator column.
5. Pump G2 is switched on. Using microvalve on the flow meter, FI1 the flow rate is
adjusted.
6. It is ensured that the wetted-wall column is wet uniformly throughout the procedure in
order to record valid data.
7. Compressor P1 is switched on and flow rate is adjusted using microvalve on the flow
meter FI2.
8. Nitrogen is allowed to flow slowly and carefully into the deoxygenator column by
opening valve V1.
During Experiment

1. The unit is setup and the column is allowed to come to steady state.
2. The tank is filled ¾ full of water before initiating the airflow.
3. The experiment is started under following conditions:
Air flow rate – 60 L/hr
Water flow rate – 4, 8, 12 and 16 L/hr
4. Step 3 is repeated with water flow rates of 120 and 180 L/hr.
5. Observations are tabulated as shown in the results.
Shut down

2
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption
1. Nitrogen supply (V1) is turned off.
2. Compressor (P1), pumps G1 and G2 are turned off.

3
 HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Results
Sample calculation for the experiment can be seen in Appendix. All relevant data are calculated using MS Excel and graphs are plotted for further
discussion of results as seen Discussion.
Table 1: Experimental Data at Different Air and Water Flow Rates

Water csat, in csat, out csat, in csat, out

Air Flow Tin Tout cin cout Patm Psat, in Psat, out Tave ν @ Tave
Flow Rate (mg/L) (mg/L) (mg/L) (mg/L)
Rate (L/h) (°C) (°C) (mg/L) (mg/L) (mmHg) (mmHg) (mmHg) (°C) (cm2/s)
(L/h) @ 1 atm @ 1 atm @ Patm @ Patm
4.0 29.10 26.90 3.91 8.95 760.00 7.65 7.95 30.21 26.58 7.65 7.95 28.00 8.416E-03
8.0 30.60 28.50 3.23 8.52 760.00 7.46 7.73 32.94 29.19 7.46 7.73 29.55 8.101E-03
60
12.0 31.40 30.20 4.56 7.96 760.00 7.36 7.51 34.47 32.19 7.36 7.51 30.80 7.896E-03
16.0 30.80 30.20 5.82 7.88 760.00 7.44 7.51 33.31 32.19 7.44 7.51 30.50 7.939E-03
4.0 30.20 29.50 4.85 8.09 760.00 7.51 7.60 32.19 30.93 7.51 7.60 29.85 8.040E-03
8.0 31.00 29.40 3.28 8.21 760.00 7.41 7.61 33.69 30.75 7.41 7.61 30.20 7.981E-03
120
12.0 31.30 30.00 4.79 7.78 760.00 7.37 7.54 34.28 31.82 7.37 7.54 30.65 7.917E-03
16.0 31.40 30.60 4.37 7.68 760.00 7.36 7.46 34.47 32.94 7.36 7.46 31.00 7.867E-03
4.0 31.00 30.30 6.63 7.89 760.00 7.41 7.50 33.69 32.38 7.41 7.50 30.65 7.917E-03
8.0 31.70 30.10 4.00 8.05 760.00 7.32 7.53 35.07 32.00 7.32 7.53 30.90 7.881E-03
180
12.0 31.70 30.50 3.66 7.64 760.00 7.32 7.48 35.07 32.75 7.32 7.48 31.10 7.853E-03
16.0 31.50 30.90 5.70 7.60 760.00 7.35 7.42 34.67 33.50 7.35 7.42 31.20 7.838E-03
 HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Table 2: Calculated Data at Different Air and Water Flow Rates

Air
Water ρ @ Wetting
Flow Na/A ∆cin ∆cout ∆cln
Flow Rate Tave Rate, W Re Na (μg/s) Kr (g/cm2.s) Kc (cm/s) Sh ln(Re) ln(Sh)
Rate (μg/s.cm2) (mg/L) (mg/L) (mg/L)
(L/h) (kg/m3) (cm2/s)
(L/h)
4.0 996.31 0.1040 49.44 5.600E+00 5.825E-03 3.74 1.00 2.07 2.799E-03 2.810E-03 10115.02 3.90 9.22
8.0 995.85 0.2080 102.72 1.176E+01 1.223E-02 4.23 0.79 2.05 5.938E-03 5.963E-03 21466.35 4.63 9.97
60
12.0 995.47 0.3121 158.10 1.133E+01 1.179E-02 2.80 0.45 1.28 9.163E-03 9.205E-03 33136.37 5.06 10.41
16.0 995.56 0.4161 209.66 9.156E+00 9.524E-03 1.62 0.37 0.84 1.126E-02 1.131E-02 40733.76 5.35 10.61
4.0 995.76 0.1040 51.75 3.600E+00 3.745E-03 2.66 0.49 1.28 2.904E-03 2.917E-03 10499.70 3.95 9.26
8.0 995.65 0.2080 104.27 1.096E+01 1.140E-02 4.13 0.60 1.83 6.208E-03 6.235E-03 22446.64 4.65 10.02
120
12.0 995.52 0.3121 157.67 9.967E+00 1.037E-02 2.58 0.24 0.99 1.047E-02 1.052E-02 37870.14 5.06 10.54
16.0 995.41 0.4161 211.56 1.471E+01 1.530E-02 2.99 0.22 1.06 1.440E-02 1.446E-02 52064.78 5.35 10.86
4.0 995.52 0.1040 52.56 1.400E+00 1.456E-03 0.78 0.39 0.56 2.580E-03 2.591E-03 9328.54 3.96 9.14
8.0 995.44 0.2080 105.59 9.000E+00 9.362E-03 3.32 0.52 1.51 6.159E-03 6.187E-03 22274.11 4.66 10.01
180
12.0 995.38 0.3121 158.96 1.327E+01 1.380E-02 3.66 0.17 1.13 1.218E-02 1.224E-02 44064.78 5.07 10.69
16.0 995.35 0.4161 212.33 8.444E+00 8.784E-03 1.65 0.18 0.66 1.328E-02 1.334E-02 48023.49 5.36 10.78
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Air Flow Rate = 60 L/h

10.80
10.60
10.40 y = 0.9783x + 5.4223
R² = 0.99725
10.20
ln(Sh)

10.00
9.80
9.60
9.40
9.20
9.00
3.00 3.50 4.00 4.50 5.00 5.50
ln(Re)

Graph 1: Air Flow rate at 60 L/h

Air Flow Rate = 120 L/h

11.00
10.80
10.60 y = 1.145x + 4.7291
10.40 R² = 0.99903
10.20
ln(Sh)

10.00
9.80
9.60
9.40
9.20
9.00
3.00 3.50 4.00 4.50 5.00 5.50
ln(Re)

Graph 2: Air FLow Rate at 120 L/h

Air Flow Rate = 180 L/h

11.00
10.80
10.60 y = 1.2395x + 4.2538
10.40 R² = 0.97747
10.20
ln(Sh)

10.00
9.80
9.60
9.40
9.20
9.00
3.00 3.50 4.00 4.50 5.00 5.50
ln(Re)

Graph 3: Air Flow Rate at 180 L/h

HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption
Discussion
Table 3: The Values of n and R2 of Different Air Flow Rates

Air Flow Rate (L/h) n R2

60 0.9783 0.9972
120 1.145 0.999
180 1.2395 0.9775

Based on the result tabulated in Table 2, the concentration flux, NA and overall mass transfer
coefficient, Kr at different water flow rates are increasing as the water flow rates increased for each
air flow rates. Moreover, as the air and water flow rates increased, the temperature will also increase.
The increase in temperature tends to separate the water molecules apart, which means the kinetic
energy of the water molecules increased. This phenomena allowed trapped oxygen molecules within
the pure water to be separated. Thus, we can concluded that the solubility of oxygen within water is
decreased.

Furthermore, by comparing the liquid phase mass transfer coefficient, Kc at constant water
flow rate but different air flow rate, it can be interpreted as it increased as the air and water flow rates
increased. However, some of the values obtained are not as expected due to the errors occurred during
experiment carried out. For instance, the gradient of the best-fit line of the graphs plotted, n is
increased as air flow rate increased. Conversely, the result of R2 values are inconsistent. By referring
the result tabulated in Table 9, we can clearly note that R 2 at 180 L/h is smaller than at 120 L/h.
Therefore, it can be concluded that experimental data was not entirely accurate.

One of the reason why the results obtained are not precise and accurate is the presence of the
rippling, which the conditions vary widely among the experimental columns. Besides, technical error
occurred in V1 for controlling the flow rates caused the results obtained are inconsistent. Therefore, it
is recommended that the experiment should be repeated in order to minimize the errors and improved
the accuracy of the result.

Conclusion
It was observed that there was a change on the density, mass transfer coefficients, flux and solubility
as a result of increasing flow rate of water and air. The flux and mass transfer coefficient increased
with the increase of water flow rates. Solubility decreased as temperature increase when air flow rate
was increased. This is because dissolved oxygen concentration also increased. However,
discrepancies of the results obtained indicates that a large set of data is needed in order to achieve
better estimation of the values of n. A longer period of time for the system to reach steady state is also
required to obtain a more accurate set of data.
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Appendix
Sample Calculation

Column diameter = 34mm

Column height = 900mm
Wetted Perimeter, P ≅ 2 πR=πD=3.1416× 3.4 cm=10.68 cm
Gas/liquid interface area, A = 10.68cm × 90cm = 961.33cm2
D = 2.5 ×10−5 cm2 /s
Wate
Csat,in Csat,out Csat,in
r Csat,out
T¿ Tout Cin Cout Patm @ Psat ,∈¿ ¿ Psat , out @ Tave ν at Tave
Flow @ 1atm @ Patm
1atm Patm
Rate
pp pp mm
L/h ℃ ℃ m m Hg
mg/L mg/L mmHg mmHg ppm ppm ℃ cm2 /s
4 29.10 26.90 3.91 8.95 760 7.65 7.95 30.21 26.58 7.65 7.95 28 8.416 ×10−3
Table 4: Data for Sample Calculation

T ¿ +T out
Average temperature, T avg ,℃ =
2
L
At water flow rate = 4 ,
h
29.10+26.90
T avg ,℃ = =28℃
2

Linear interpolation method is used to find C sat ,∈¿¿ by extracted from Table 1 in lab manual.
At T ¿=29.10 ℃ ,
Table 5: Interpolation C sat ,∈¿¿ at T ¿=29.10 ℃

T℃ mg
C sat ,∈¿¿ @ 1 atm,
L
29 7.66
29.10 C sat ,∈¿¿
30 7.54

Linear interpolation;
30−29 7.54−7.66
=
30−29.10 7.54−C sat ,∈¿ ¿
C mg
sat ,∈¿=7.65 ¿
L

*** The calculation is repeated to find C sat ,∈¿¿C sat,out ¿ at different temperature for each flow rate.
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Linear interpolation method is used to find Psat ,∈¿ ¿ by extracted from Table 3 in lab manual.
At T ¿=29.10 ℃
1 atm = 101325 Pa = 760 mmHg
Table 6: Interpolation Psat ,∈¿ ¿ at T ¿=29.10 ℃

T℃ Psat , Pa
29 4004
29.10 Psat ,∈¿ ¿
30 4242

Linear interpolation;
30−29 4242−4004
=
30−29.10 4242−Psat ,∈¿ ¿
P 760mmHg
sat ,∈¿=4027.8 Pa × =30.21 mmHg¿
101325 Pa

*** The calculation is repeated to find Psat ,∈¿ ¿P sat, out ¿ at different temperature for each flow rate.
Linear interpolation method is used to find kinematic viscosity , ν by extracted from Table 2 in
lab manual.
At T avg=28 ℃
Table 7: Interpolation kinematic viscosity , ν At T avg=28 ℃

m2
T℃ Kinematic viscosity , ν
s
20 1.004 ×10−6
28 ν
30 0.801 ×10−6

Linear interpolation;
30−20 0.801 ×10−6 −1.004 ×10−6
=
30−28 0.801 ×10−6−ν
m 2 10000 cm2 2
−6 −3 cm
ν=0.8416 ×10 × =8.416 × 10
s 1 m2 s
*** The calculation is repeated to find ν at different average temperature for each flow rate.
 HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption
Table 8: Calculated Data at 4.0 L/h of Water Flow Rate

Water Wetting
ρ @ Tave Na/A ∆cin ∆cout ∆cln
Flow Rate Rate, W Re Na (μg/s) Kr (g/cm2.s) Kc (cm/s) Sh ln(Re) ln(Sh)
(kg/m3) (μg/s.cm2) (mg/L) (mg/L) (mg/L)
(L/h) (cm2/s)
8.0 995.85 0.2080 102.72 1.176E+01 1.223E-02 4.23 0.79 2.05 5.938E-03 5.963E-03 21466.35 4.63 9.97

Linear interpolation method is used to find Density , ρ by extracted from Table 3 in lab manual.
At T avg=29.55 ℃ ,
Table 9: Interpolation Density , ρ at T avg=29.55 ℃

T (°C) ρ (kg/m3)
29.00 996.02
29.55 ρ
30.00 995.71

Linear interpolation;
30−29 995.71−996.02
=
30−29.55 995.71−ρ
kg
ρ=995.85
m3
*** The calculation is repeated to find ρ at different average temperature for each flow rate.
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption

Water Flow Rate

¿
Wetted Perimeter

L cm3 1 h
4 ×1000 ×
h L 3600 s
¿
10.68 cm
cm2
¿ 0.1040
L

Reynolds Number , ℜ cm2

4 ×0.1040
s
¿
cm2
8.416 ×10−3
s
¿ 49.43
Absorption Rate , Na
¿ Water Flow Rate × ( c out −c ¿ )
L mg 1 h
¿4 × ( 8.95−3.91 ) ×
h L 3600 s
mg
¿ 0.0056
s
μg
¿ 5.6
s
μg
5.6
s
Na ¿
961.33 cm2
A
μg
¿ 0.00583
s . cm2
∆ C¿ ¿ C sat ,∈¿@ P −C ¿ ¿
atm

¿ 7.65−3.91
mg
¿ 3.74
L
∆ C out ¿|C sat ,0 ut @ P atm−C out|
¿|7.95−8.95|
mg
¿1
L
∆ c ln ( ∆ C ¿−∆ Cout )
¿
ln
( ∆C∆ C )
¿

out
HEC 4722 Process Mass Transfer Group A
Lab 3 Gas Absorption
(3.74−1)
¿
3.74
ln( ) 1
mg
¿ 2.077
L
Kr Na
A
¿
∆ c ln
g g
0.00583× 10−6 ×996.31
s . cm 2
L
¿
g
2.077 ×10−3
L
g
¿ 2.796 ×10−3
s .cm 2
Kc Kr
¿
ρ
g
2.796× 10−3
s . cm2
¿
g
0.99631 3
cm
cm
¿ 2.807 ×10−3
s
Sh K c ×Column Height
¿
D
cm
2.807× 10−3 ×90 cm
s
¿
cm2
2.5 E−5
s
¿ 10105