Document4

1.1 Abstract

The Gas Absorption experiment was conducted in order to determine the loading and flooding

points in the column and to determine the air pressure drop across the column as a function of

airflow for different water flow rates through the column. The result obtained is to be compared

between theoretical values that has been calculated. The experiment was run three times with

different water flow rate which are 1.0 L/min, 2.0 L/min and 3.0 L/min. For every water flow

rate, it was run for different air flow rate of 20 L/min, 40 L/min, 60 L/min, 80 L/min, 100 L/min,

120 L/min, 140 L/min, 160 L/min and 180 L/min. Graph of pressure drop against the air flow

rate was plotted and it showed an increasing pattern for all three water flow rates. The flooding

point was recorded during the water flow rate of 3.0 L/min and air flow rate of 100 L/min. The

value of pressure drop taken was 14 mm H2O with 3 mm H2O as compared in Appendix.

Basically the pressure drop is increasing when the air flow rate increased. The flooding

happened when the air pressure from the bottom is too high and pushed the water up. The

percentage different between the pressure drop gained from Appendix and the one

recorded is 78.57%.

1.2 Introduction

Gas absorption is a process in which one or more soluble components which is solutes are

removed from a gas phase by contact with a liquid phase which is solvent into which the

components of interest dissolve. Basically, the absorbed gas is removed from the solvent and the

solvent liquid stream is returned to the system. This solvent-recovery process is called stripping.

Stripping also happened when volatile components have to be removed from a liquid mixture.

The stripping agent can be either a gas which is air or a superheated vapour which is superheated

steam. Absorption and stripping employ special contactors for bringing gas and liquid phase into

intimate contact. (Coca, 2007)

In most common unit gas absorption, the solvent will enters the top of the absorber or stripper

and will flows downwards, counter current to the rising gas stream. The two phases mix and will

contact one another and the solute is transferred from the gas phase to the solvent (Coca, 2007).

A schematic diagram of an absorption-stripping process is shown as in Figure 1:

Figure 1: Typical absorption-stripping process with recycle of solvent (a) absorber, (b) stripper

There are two types of absorption processes, physical absorption and chemical absorption. The

process is dependent on whether a chemical reaction occurs between the solute and solvent that

act as absorbent. The process is classified as physical absorption when no chemical reaction

occurs between the solute and the absorbent and chemical absorption occurs when there is

irreversible and rapid neutralization reaction in the liquid phase (Gas Absorption & Desorption).

The absorption column unit model BP: 751-B that been used in this experiment is meant to

demonstrate the absorption of air into water in a packed column where the gas and liquid mixture

will counter-currently flow among the column packing. It is also designed to operate at

atmospheric pressure in continuous operation.

A common apparatus used in the absorption column and other operation is the packed tower.

This device consists of a tower equipped with a gas inlet at the bottom as a distributing space, a

liquid inlet and distributor at the top. The gas and liquid outlet are at the top and bottom

respectively and tower packing that use as a supported mass of inert solid shapes. The packing is

used to increase the surface contact area between the gas and liquid absorbent.

1.3 Aims

The objectives of this experiment is:

i. To determine the loading and flooding points in the column.

ii. To determine the air pressure drop across the column as a function of airflow for different

water flow rates through the column.

1.4 Theory

In an absorption process, two immiscible phases which are liquid and gas are present in which

the solute will diffuse from one phase to the other through the interface between the phases. For

a solute A to diffuse from the gas phase (V) into the liquid phase (L), it must pass through phase

V first, next the interface and then into the phase L in series.

A concentration gradient has to exist to allow the mass transfer to take place through resistances

in each phase. The concentration in the bulk gas phase, y(L) decreases to y(i) at the interface,

while the liquid concentration starts at x(i) at the interface and decreases to x(L) the bulk liquid

phase concentration.

There are two types of diffusions in an absorption process:

I. Equimolar counter diffusion – two components diffusing across the interface, one from

the gas to liquid phase, and the other are from the liquid to gas phase.

II. Diffusion through stagnant (Non-diffusing phase) – only one component diffuses across

the interface through stagnant gas and liquid phases.

This experiment required to plot graph of pressure drop against air flow rate in graph. The flow

parameter shows the ratio of liquid kinetic energy to vapour kinetic energy and parameter of K 4

or y-axis needs and x-axis or FLV can be calculated by using these formulae:

G y2 FP μx

0.1

gc ( ρ x−ρ y) ρ y

Gx

G y √ ρ y

ρx−ρ yGas absorption is a process where mixture of gas is in contact with

liquid and becomes dissolve. Therefore, there is mass transfer occurs in the component that

changes from gas phase to liquid phase. The solutes are absorbed by liquid. Inside this

experiment, only the mass transfer between air and liquid are concerned. Gas absorption is

widely use in industries to control the air pollution and to separate acidic impurities out of mixed

gas streams. The pressure drop values are observed from the manometer. The graph of pressure

correlation for different flow rate of water is plotted in order to find the relationship between K4

and FLV. The steps on how to obtaine K4 and FLV is shown below:

Density of air, ρG = 1.175 kg/m3

Density of water, ρL = 996 kg/m3

Column diameter, Dc = 80 mm

Area of packed diameter, Ac=π4

D2

Packing Factor: Fp = 900 m-1

Water viscosity, µwater = 0.001 Ns/m2

Theoretical Flooding Point

1. Gy must be in m3/h

2. To calculate gas flow rate, GG (kg/m2s)

GG=G y × ρ

Ac

3. To calculate capacity parameter, K4,

K4=

13.1 (GG )2 F p( μL

ρL)

0.1

ρG ( ρL−ρG )

4. To calculate liquid flow rate, GL (kg/m2) (1 LPM, 2 LPM, 3 LPM)

GL=G × ρ

Ac

5. To calculate flow parameter, FLV (1 LPM)

FLV=GL

GG(√ ρG

ρL)

Where:

G y = Air flow rate (m3/h)

1.5 Apparatus

Figure 1: Gas Absorption Column Unit (Model: BP 751-B)

1.6 Procedure

General start-up Procedures

1. All the valves were ensured closed except the ventilation valve V13.2. All the gas connections were checked whether they are properly fitted.3. The valve on the compressed air supply line was opened. The supply

pressure was set between 2 to 3 bar by turning the regulator knob clockwise.

4. The shut-off valve on the CO2 gas cylinder was opened. The CO2 cylinder was checked whether the pressure is sufficient.

5. The power for the control panel was turned on.

Experiment : Hydrodynamics of a Packed Column (Wet Column Pressure Drop)

1. The general start-up procedures were performed as described above.2. The receiving vessel B2 was filled through the charge port with 50 L of

water by opening valve V3 and V5.3. Valve V3 was closed.4. Valve V10 and V9 were opened slightly. The flow of water was

observed from vessel B1 through pump P1.5. Pump P1 was switched on, then slowly opened and valve V11 was

adjusted to give water flow rate of arrounf 1 L/min. The water was allowed to enter the top column K1, flew down the column and accumulated at the bottom until it overflows back into vessel B1.

6. Valve V11 was opened and adjusted to give water flow rate of 0.5 L/min into column K1.

7. Valve V1 was opened and adjusted to give an air flow rate of 40 L/min into column K1.

8. The liquid and gas flow were observed in the column K1 and the pressure drop across the column at dPT-201 was recorded.

9. Steps 6 to 7 were repeated with different values of air flow rate, each time increasing by 40 L/min while maintaining the same water flow rate.

10. Steps 5 to 8 were repeated with different values of water flow rate, each time increasing by 0.5 L/min by adjusting valve V11.

General Shut-Down Procedures

1. Pump P1 was switched off.2. Valve V1, V2 and V12 were closed.3. The valve on the compressed air supply line was closed and the supply

pressure was exhausted by turning the regulator knob counter-clockwise all the way.

4. The shut-off valve was closed on the CO2 gas cylinder.5. All the liquid in the column K1 was drained by opening valve V4 and

V5.6. All the liquid from the receiving vessels B1 and B2 were drained by

opening valves V7 and V8.7. All the liquid from the pump P1 was drained by opening valve V10.8. The power for the control panel was turned off.

1.7 Results

flow

rate(L/min)

Pressure drop (mm H2O)

Air

Water

20 40 60 80 100 120 140 160 180

1.0 0 1 2 4 7 11 17 26 48

2.0 19 24 29 33 43 - - - -

3.0 11 19 25 49 - - - - -

Table 1: Pressure Drop for Wet column

20 40 60 80 100 120 140 160 1800

10

20

30

40

50

60

70

80

90

100

Graph of Pressure drop against air flow rate

3 LPM2 LPM1 LPM

Graph 1: Graph Of Pressure Drop Against Air Flow Rate

Flow rate

(L/min)

1.0 2.0 3.0 Water

Air

Gas

Flow

rate

Log

Gas

Flow

rate

Pressure

drop

(mmH2O)

Log

Pressure

drop

(mmH2O)

Pressure

drop

(mmH2O)

Log

Pressure

drop

(mmH2O)

Pressure

drop

(mmH2O)

Log

Pressure

drop

(mmH2O)

20 1.301 0 - 19 1.279 11 1.041

40 1.602 1 0 24 1.380 19 1.279

60 1.778 2 0.301 29 1.462 25 1.398

80 1.903 4 0.602 33 1.519 49 1.690

100 2.00 7 0.845 43 1.633 - -

120 2.079 11 1.041 - - - -

140 2.146 17 1.230 - - - -

160 2.204 26 1.415 - - - -

180 2.255 48 1.681 - - - -

Table 2: Log Gas Flow rate and Log Pressure drop

Air

Flow

rate

(L/min)

Air

Flow

rate

(m3/h)

GG

(kg/ms2)

K4

(y-axis)

FLV (1

LPM)

FLV (2

LPM)

FLV (3

LPM)

Pressure drop

correlated in mm H20

(1LPM) (2LPM)

(3LPM)

20 1.2 0.0779 0.0154 1.454 2.912 4.362 1.52 5.08 12.740 2.4 0.156 0.062 0.727 1.456 2.181 5.08 10.16 25.4

60 3.6 0.234 0.139 0.484 0.971 1.454 8.89 25.4 38.180 4.8 0.311 0.245 0.364 0.729 1.093 12.7 40.64 50.8

100 6.0 0.389 0.383 0.292 0.584 0.874 25.4 45.72 -

120 7.2 0.467 0.553 0.243 0.468 0.728 40.64 - -

140 8.4 0.545 0.753 0.208 0.416 0.624 43.18 - -160 9.6 0.623 0.989 0.182 0.365 0.546 50.80 - -180 10.8 0.701 1.245 0.162 0.324 0.485 55.89 - -

Table 3: Theoretical Flooding Point

1.301 1.602 1.778 1.903 2 2.079 2.146 2.204 2.2550

0.5

1

1.5

2

2.5

3

3.5

4

4.5

Graph of Log Pressure Drop against Log Gas Flow Rate

3 LPM2 LPM1 LPM

Log Gas Flow Rate, Gy

Log

Pres

sure

Dro

p, m

mH2

O

Graph 2: Graph of Log Pressure Drop against Log Gas Flow Rate

0.0154 0.062 0.139 0.245 0.383 0.553 0.753 0.989 1.2450

1

2

3

4

5

6

7

8

9

10

Graph of K4 against FLV

3 LPM2 LPM1 LPM

Graph 3: Graph of K4 against FLV

1.8 Calculations

Information given :

Density of air, ρair = 1.175 kg/m3

Density of water, ρwater = 996 kg/m3

Column diameter, Dc = 80 mm

Area of packed diameter,

Ac=π4

D2= π4

(0.08m )2=5.027 ×10−3 m2

Packing Factor: Fp = 900 m-1

Water viscosity, µwater = 0.001 Ns/m2

LIQUID

Calculate liquid flow rate, GL (kg/m2) (1 LPM)

GL=G × ρ

Ac=

1Lmin

× 1min60 s

× 1m3

1000l× 996 kg

m3

5.027 ×10−3m2 ¿3.302 kg

m2 . s


GL=G × ρ

Ac=

2Lmin

× 1min60 s

× 1m3

1000l× 996 kg

m3

5.027 ×10−3m2 =6.614 kgm2 . s


GL=G × ρ

Ac=

3Lmin

× 1min60 s

× 1m3

1000l× 996 kg

m3

5.027 ×10−3m2 =9.907 kgm2 . s

Water Flow Rate (L/min)GL (

kgm2 . s

)

1.0 3.302

2.0 6.614

3.0 9.907

Table

Theoretical Flooding Point for 20 L/min

Gy = 20 Lmin

× 1m3

1000 L× 60min

1 h

= 1.2 m3 /h

Calculate gas flow rate, GG (kg/m2s)

GG=G y × ρ

Ac

¿

1.2 m3

h× 1 h

3600 s× 1.175 kg

m3

5.027 ×10−3 m2 =0.0779 kgm2 . s

Calculate capacity parameter, K4 , y-axis

K 4=

13.1 (GG )2 F p( μL

ρL )0.1

ρG ( ρL−ρG )

¿13.1( 0.0779 kg

m2 s )2

(900 m−1 )( 0.001 N . s /m2

996 kg/m3 )0.1

( 1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.0154

Calculate flow parameter, FLV (1 LPM)

FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.0779 kgm2 . s

(√ 1.175 kg /m3

996 kg/m3 )=1.454


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2. s

0.0779 kgm2 . s

(√ 1.175 kg /m3

996 kg/m3 )=2.912


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2 . s

0.0779 kgm2 . s

(√ 1.175 kg /m3

996 kg/m3 )=4.362


Gy =40 Lmin

× 1m3

1000 L× 60 min

1h=2.4 m3/h


GG=G y × ρ

Ac

GG=

2.4 m3

h× 1 h

3600 s× 1.175 kg

m3

5.027 ×10−3 m2 =0.156 kgm2 . s

Calculate capacity parameter, K4,

K 4=13.1( 0.156 kg

m2 s )2

(900 m−1 )( 0.001 N . s/m2

996 kg /m3 )0.1

(1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.062


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.156 kgm2. s

(√ 1.175 kg/m3

996 kg/m3 )=0.727


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2. s

0.156 kgm2. s

(√ 1.175 kg/m3

996 kg /m3 )=1.456


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2 . s

0.156 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=2.181


Gy = 60 Lmin

× 1 m3

1000 L× 60min

1h=3.6 m3/h


GG=

3.6 m3

h× 1h

3600 s× 1.175 kg

m3

5.027 ×10−3m2 =0.234 kgm2 . s


K 4=13.1( 0.234 kg

m2 s )2

(900 m−1 )( 0.001 N . s /m2

996 kg /m3 )0.1

(1.175 kg/m3 ) (996 kg /m3−1.175 kg /m3 )=0.139


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.234 kgm2. s

(√ 1.175 kg/m3

996 kg /m3 )=0.484


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2 . s

0.234 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=0.971


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2. s

0.234 kgm2. s

(√ 1.175 kg/m3

996kg /m3 )=1.454


Gy = 80 Lmin

× 1 m3

1000 L× 60 min

1h=4.8 m3/h


GG=

4.8 m3

h× 1 h

3600 s× 1.175 kg

m3

5.027 ×10−3 m2 =0.311 kgm2 . s


K 4=13.1( 0.156 kg

m2 s )2

(900 m−1 )( 0.001 N . s/m2

996 kg /m3 )0.1

(1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.245


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.311 kgm2 . s

(√ 1.175 kg /m3

996 kg /m3 )=0.364


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2 . s

0.311 kgm2 . s

(√ 1.175 kg/m3

996kg /m3 )=0.729


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2 . s

0.311 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=1.093


Gy = 100 Lmin

× 1 m3

1000 L× 60min

1h=6.0 m3/h


GG=

6.0 m3

h× 1 h

3600 s× 1.175 kg

m3

5.027 ×10−3m2 =0.389 kgm2 . s


K 4=13.1( 0.156 kg

m2 s )2

(900 m−1 )( 0.001 N . s/m2

996 kg /m3 )0.1

(1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.383


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.389 kgm2 . s

(√ 1.175 kg /m3

996 kg/m3 )=0.292


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2 . s

0.389 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=0.584


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2 . s

0.389 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=0.874


Gy = 120 Lmin

× 1 m3

1000 L× 60 min

1 h=7.2 m3/h


GG=

7.2 m3

h× 1 h

3600 s× 1.175 kg

m3

5.027 ×10−3 m2 =0.467 kgm2 . s


K 4=13.1( 0.467 kg

m2 s )2

(900 m−1 )( 0.001 N . s/m2

996 kg /m3 )0.1

(1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.553


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.467 kgm2 . s

(√ 1.175 kg/m3

996 kg/m3 )=0.243


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2. s

0.467 kgm2. s

(√ 1.175 kg/m3

996 kg /m3 )=0.486


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2 . s

0.467 kgm2 . s

(√ 1.175 kg/m3

996kg /m3 )=0.728


Gy = 140 Lmin

× 1 m3

1000 L× 60min

1h=8.4 m3 /h


GG=

8.4 m3

h× 1h

3600 s× 1.175 kg

m3

5.027 ×10−3 m2 =0.545 kgm2 . s


K 4=13.1( 0.545 kg

m2 s )2

(900 m−1 )( 0.001 N .s /m2

996 kg/m3 )0.1

( 1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.753


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.545 kgm2 . s

(√ 1.175 kg /m3

996 kg/m3 )=0.208


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2. s

0.545 kgm2. s

(√ 1.175 kg/m3

996 kg /m3 )=0.416


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2 . s

0.545 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=0.624


Gy = 160 Lmin

× 1 m3

1000 L× 60min

1h=9.6 m3/h


GG=

9.6 m3

h× 1 h

3600 s× 1.175 kg

m3

5.027 ×10−3 m2 =0.623 kgm2 . s


K 4=13.1( 0.623 kg

m2 s )2

(900 m−1 )( 0.001 N .s /m2

996 kg/m3 )0.1

( 1.175 kg/m3 ) (996 kg /m3−1.175 kg/m3 )=0.989


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.623 kgm2 . s

(√ 1.175 kg /m3

996 kg/m3 )=0.182


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2 . s

0.623 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=0.365


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2. s

0.623 kgm2. s

(√ 1.175 kg/m3

996 kg /m3 )=0.546


Gy = 180 Lmin

× 1 m3

1000 L× 60min

1h=10.8 m3/h


GG=

10.8 m3

h× 1h

3600 s× 1.175 kg

m3

5.027 ×10−3m2 =0.701 kgm2. s


K 4=13.1( 0.701 kg

m2 s )2

( 900 m−1 )( 0.001 N .s /m2

996 kg/m3 )0.1

(1.175 kg /m3 ) ( 996 kg /m3−1.175 kg/m3 )=1.245


FLV=GL

GG(√ ρG

ρL)=

3.302 kgm2 . s

0.701 kgm2 . s

(√ 1.175 kg /m3

996 kg /m3 )=0.162


FLV=GL

GG(√ ρG

ρL)=

6.614 kgm2 . s

0.701 kgm2 . s

(√ 1.175 k g /m3

996 kg/m3 )=0.324


FLV=GL

GG(√ ρG

ρL)=

9.907 kgm2. s

0.701 kgm2 . s

(√ 1.175 kg/m3

996 kg /m3 )=0.485

PERCENTAGE ERROR %

1LPM

Total pressure drop = 116 mm H20

Percentage error , %= theoritical value−experimental valuetheoritical value

× 100 %

¿ 244.1−116244.1

×100 %=52.48 %

2LPM



× 100 %

¿ 127−148127

×100 %=−16.54 %

3LPM



× 100 %

¿ 127−104127

×100 %=18.11%

1.9 Discussion

This experiment was conducted to determine the pressure drop across different flow rates of

water which is 1LPM, 2LPM and 3LPM. Pressure drop occurs in the packed column due to the

difference in pressure at the top and bottom of the column that caused by the interface between

the liquid and gas flow streams in the packed column.

In this experiment, the data was tabulated based on the necessary formula given. As shown in the

calculation part, the pressure drop based on the equipment was stated.

By referring to the data collected in Table 1: Pressure Drop for Wet Column, graph of

pressure drop vs flow rate are plotted as shown in Figure 3. From the graph, we can conclude

that as the flow rate of air increase, the pressure drop will also be increased. The pattern of

pressure drop in this graph increased proportionally with the air flow rate. This is because the gas

started to hinder the liquid down flow and local accumulation of liquid started to appear in the

packing.

Percentage error of pressure drop in 1LPM is 48%, 2LPM is 46% and 3LPM is 5.08%.

This is due to overflow during experiment is carried on. It also might be due to minor leaking

when the experiment is being carried out. Minor leaking will affect the flow rate of both water

and air thus affecting the pressure drop. When the gas flow rate increased, pressure drop

increased and some of the water will trapped in packing. Later, the water from bottom will

increase until the highest level and this will results in flooding.

Flooding happened when the gas velocity is very high and it does not allow the flow of liquid

from the top of tower and flooding occurs on top of it. Based on theory, the best gas velocity is it

should be half of the flooding velocity. Flooding is a major concern in the industrial process

where it is a phenomenon by which gas moving in one direction in the packed column entrains

liquid moving in the opposite direction in the packed column. Based on this experiment, at the

point where the flow rate of air is 100 L/min and water flow rate of 3.0 L/min the pressure drop

has achieved its flooding point. At that point, the water level has reached its limit on the warning

line on the instrument and water flow out from the top of the tower.

In industrial, flooding is undesirable because it can cause a large pressure drop across the packed

column as well as other effects that can effect the performance and stability of the absorption

process. Apart of that, many parameters need to be considered for efficiency and to avoid

flooding problem in the design of the absorption packed column.

Unfortunately, there are few aspects need to be considered to get a better result for this

experiment. Before start the experiment, set properly the packed bed to avoid from equipment

error. For the best result also, before conducting the experiment, checked the condition of the

compressor first to ensure that the compressor operates normally. The air valve also should be

checked to make sure that it can adjusted to desired flow rates. This is because the result cannot

be recorded if we can’t get the right air flow rates.

1.10 Conclusion

In conclusion, the objective of this experiment was achieved. The air pressure drop across the

column for different water flow rates and the loading and flooding point in the column have been

determined. Based on the result, it can be said that as the gas flow increase at constant water flow

rates, the pressure drop will increase as well. The loading and flooding point was determined

whereby the flooding point of the water flow rates is 3.0L/MIN at 100 LPM respectively. The

importance of knowing the flooding point of the equipment been used is for user to notify the

limit at which level the equipment will faced up the flooding point. From this experiment, it can

be concluded that the air pressure drop across the column increases as the air flow rate and water

flow rate increases through the column.

1.11 Recommendations

In my opinion, there are some recommendations that should be taken account into to ensure the

experiment become more accurate. We should constantly checked the valve controlling the level

of water flowing back to the water reservoir to get a better reading. In order to avoid from the air

being trapped in line, the level of water must be higher than the bottom of the reservoir. Besides

that, before using all the column, make sure that all the valves are closed so that the experiment

will run smoothly and make sure that the gas and liquid flow rates were constant at that

particular flow rates. The sample also must be simultaneously collect from both inlet and outlet

of the packed column. When starting up the system, always use low initial air and water

velocities. Be sure that the recycle valve to the sump pump is always at least partially open to

prevent from build-up of liquid and flooding, an extension has been added to the top of the

column to help prevent spillage of caustic. In order to avoid a parallax error, make sure that the

eyes are in parallel level while taking the reading.

1.12 Reference

Absorption. (n.d.). Gas Absorption & Desorption. Retrieved on 11th of November fromhttp://www.separationprocesses.com/Absorption/GA_Chp03.htm

Coca, J. (2007). Mass Transfer Operations: Absorption and Extraction. Spain: Eolss.

Geankoplis, C. J. (2003). Transport Processes and Separation Process Principle. Fourth edition,

page 659.

Yunus A. Cengel et. al., Fluid Mechanics Fundamentals and Applications, 2nd Edition, McGraw

Hill.

http://www.separationprocesses.com/Absorption/GA_Chp03.htm

1.13 Appendix

UNIVERSITI TEKNOLOGI MARAFAKULTI KEJURUTERAAN KIMIA

CHEMICAL ENGINEERING LABORATORY 1 (CPE 465)

NAME : NUR IZZATI BINTI AHMAD TARHIZIGROUP : EH200 2AEXPERIMENT : FLUID MIXINGDATE : 28 APRIL 2015PROG/CODE : EH220SUBMIT TO : MADAM NURUL DIYANAH

No Title Allocated Marks (%) Marks1 Abstract 52 Introduction 53 Objectives 54 Theory 55 Procedures/Methodology 106 Apparatus 57 Results 108 Calculation 109 Discussion 20

10 Conclusion 1011 Recommendations 512 References 513 Appendices 5

TOTAL 100

Remarks:

Checked by: Rechecked by:

Date: Date:

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