Gas Absorption

xzNAME STUDENT NO GROUP EXPERIMENT DATE PERFORMED SEMESTER PROGRAMME / CODE 554 ) SUBMIT TO NO. 1 2 3 4 5 6 7 8 9 10 11 12 13

: SHEH MUHAMMAD AFNAN BIN SEH HANAFI : 2013210382 : EH2214A : GAS ABSORPTION : 11/03/2015 : 4 : PROCESS ENGINEERING LABORATORY ( CPE : MISS HABSAH ALWI

TITLE ABSTRACT / SUMMARY INTRODUCTION AIMS THEORY APPARATUS METHODOLOGY / PROCEDURE RESULTS CALCULATIONS DISCUSSION CONCLUSION RECOMMENDATIONS REFERENCE APPENDIX TOTAL MARKS

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TABLE CONTENT

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Number 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13.

Title Abstract/Summary Introduction Aims Theory Apparatus Methodology/Procedure Results Calculations Discussion Conclusion Recommendations Reference Appendix

Page 2 3 3 4–5 6 7–8 9 – 11 12 - 22 23 24 24 25 26

ABSTRACT Gas absorption is a process in which a gaseous mixture is brought into contact with a liquid and during this contact a component is transferred between the gas stream and the liquid stream. The gas may be bubbled through the liquid, or it may pass over streams of the liquid, arranged to provide a large surface through which the mass transfer can occur. The liquid film can flow down the sides of columns or over packing, or it can cascade from one tray to another with the liquid falling and the gas rising in the counter flow. The gas, or components of it, either dissolves in the liquid (absorption) or extracts a volatile component from the liquid (desorption). In addition, there is the aim that should be achieved at the end of the experiment which is to examine the air pressure drop across the column as a function of

1

air flow for different water flow rates through the column. In packed column, air is fed into the bottom and water is transferred to the top of the column either from feed vessel B1 using the centrifugal pump, P1. The pressure drop is recorded when the liquid flow rate is set to 1.0 L/min until 3.0 L/min. The gas flow rate starts from 20 L/min until 180 L/min with 10.0L/min of intervals. How fast the liquid can flow down with no vapor flowing upwards and the rate at which the vapor is trying to flow upwards is the actual flooding point.

INTRODUCTION Gas absorption is a mass transfer process in which a vapor solute A in a gas mixture is absorbed by means of a liquid in which the solute more or less soluble. The gas mixture consists mainly of an inert gas and the soluble. The liquid also is primarily in the gas phase; that is, its vaporization into the gas phase is relatively slight. A typical example is absorption of the solute ammonia from an air-ammonia mixture by water. Subsequently, the solute is recovered from the solution by distillation. A common apparatus used in gas absorption and certain other operations is the packed tower. The device consists of a cylindrical column, or tower, equipped with a gas inlet an distributing space at the bottom, a liquid inlet and distributor at the top, gas and liquid outlet at the top and bottom, respectively and a supported mass of inert solid shapes, called tower packing. 2

In given packed tower with a given type and size of packing and with defined flow of liquid, there is an upper limit to the rate of gas flow, called the flooding velocity. Above this gas velocity the tower cannot operate. At the flow rate called the loading point, the gas start to hander the liquid down flow, and local accumulations or pools of liquid start to appear in the packing.

AIMS 1. To examine the air pressure drop across the column as a function of air flow for different water flow rates through the column. 2. To determine the loading and the flooding points in the column.

THEORY 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 K4 or y-axis needs and x-axis or FLV can be calculated by using these formulae: G2y F P μ0.1 Gx x g c ( ρ x −ρ y ) ρ y G y

√

ρy ρx −ρ y

Gas 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 3

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 K 4 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,

A c=

π 2 D 4

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) G G=

3.

Gy × ρ Ac

To calculate capacity parameter, K4,

4

2

13.1 ( GG ) F p K4=

μL ρL

0.1

( )

ρG ( ρ L− 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: Gy

= Air flow rate (m3/h)

APPARATUS

SOLTEQ-QVF Absorption column (Model: BP 751-B)

5

PROCEDURE General start-up Procedures 1. All the valves were ensured closed except the ventilation valve V13. 6

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. 7

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.

RESULTS

8

Flow

Pressure Drop

rate

(mm

H2O ¿

(L/min) air 20

40

60

80

100

120

140

160

180

2 2 2

6 0 7

7 3 13

9 9 39

10 30 -

13 53 -

21 -

58 -

water 1.0 2.0 3.0

0 8 0

Table 1: Pressure Drop for Wet column

Flow rate (L/min)

1.0

2.0

3.0

Air Water Gas

Log

Pressure

Log

Pressure

Log

Pressure

Log

Flow

Gas

drop

Pressure

drop

Pressure

drop

Pressure

rate

Flow

(mmH2O)

drop

(mmH2O

drop

(mmH2O

drop

(mmH2O

)

(mmH2O

)

(mmH2O

rate 20 40 60 80 100 120 140 160 180

1.301 1.602 1.778 1.903 2.00 2.079 2.146 2.204 2.255

) 8 0.301 2 0.778 0 0.845 3 0.954 9 1 30 1.114 53 1.322 1.763 Table 2: Log Gas Flow rate and Log

0 2 6 7 9 10 13 21 58

9

) 0.903 0 0.301 2 7 0.477 13 0.954 39 1.477 1.724 Pressure drop

) 0.301 0.845 1.114 1.591 -

Air

Air

GG

K4

Flow

Flow

(kg/ms2

rate

rate

)

(L/min)

(m3/h)

FLV (1

FLV (2

FLV (3

Pressure drop

LPM)

LPM)

LPM)

correlated in mm H20 (1LPM) (2LPM) (3LPM)

20 40

1.2 2.4

0.0779 0.156

0.0154 0.062

1.454 0.727

2.912 1.456

4.362 2.181

60 80

3.6 4.8

0.234 0.311

0.139 0.245

0.484 0.364

0.969 0.729

1.452 1.093

1.52 5.08

5.08 10.1

12.7 25.4

8.89 12.7

6 25.4 40.6

38.1 50.8

25.4

4 45.7

53.3 4

100

6.0

0.389

0.383

0.291

0.583

0.874

120

7.2

0.467

0.553

0.243

0.486

-

40.6

2 50.8

140

8.4

0.545

0.753

0.208

0.416

-

4 43.1

55.8

-

160

9.6

0.623

0.984

0.182

-

-

8 50.8

-

-

-

0 55.8

-

-

180

10.8

0.701

1.245

0.162

-

9 Table 3: Theoretical Flooding Point Log Pressure drop correlated in mm H 20 (1LPM) (2LPM) (3LPM) 0.18 0.71 0.71 0.95 1.10 1.40 1.61 1.64 1.71 1.75 Table 4 : Log Pressure

1.10

1.00 1.40 1.40 1.58 1.61 1.71 1.66 1.73 1.71 1.75 drop correlated in mm H20

10

-

Graph of Log Pressure Drop against Log Gas Flow Rate 4 3.5 3 2.5 2 1.5 1 0.5 0

Log Pressure Drop, mmH2O

3 LPM 2 LPM 1 LPM

Log Gas Flow Rate, Gy

Figure 1: Graph of Log Pressure Drop against Log Gas Flow Rate

Graph of log correlated pressure drop vs log gas flow rate 2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0

log correlated pressure drop

11

1 LPM 2 LPM 3 LPM

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

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,

A c=

π 2 π D = ( 0.08 m )2=5.027 ×10−3 m 2 4 4

Packing Factor: Fp = 900 m-1 Water viscosity, µwater = 0.001 Ns/m2 Theoretical Flooding Point for 20 L/min 1. Gy = 20 L/min 3

= 2.

G G=

20 L 1m 60 min × × =1.2 m3 /h min 1000 L 1h

Calculate gas flow rate, GG (kg/m2s)

Gy × ρ Ac 3

1.2 m 1h 1.175 kg × × 3 h 3600 s m kg ¿ =0.0779 2 −3 2 5.027 ×10 m m .s

3.

Calculate capacity parameter, K4, 12

2

13.1 ( GG ) F p K4=

¿

μL L ρ

0.1

( )

ρG ( ρ L− ρG )

2 0.0779 kg 2 ( −1 0.001 N . s /m ) 13.1 900 m m2 s 996 kg /m3

(

(

)

0.1

)

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

=0.0154

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

1 L 1 min 1m3 996 kg × × × min 60 s 1000l G× ρ m3 GL = = Ac 5.027 ×10−3 m2

¿ 3.302

kg m2 . s

5. Calculate liquid flow rate, GL (kg/m2) (2 LPM)

2 L 1 min 1m3 996 kg × × × G× ρ min 60 s 1000l m3 kg GL = = =6.614 2 −3 2 Ac 5.027 ×10 m m .s 6. Calculate liquid flow rate, GL (kg/m2) (3 LPM) 3

3 L 1 min 1m 996 kg × × × G× ρ min 60 s 1000l m3 kg GL = = =9.907 2 −3 2 Ac 5.027 ×10 m m .s 7. Calculate flow parameter, FLV (1 LPM)

F LV =

GL GG

kg ρG m2 . s = ρL kg 0.0779 2 m .s

(√ )

3.302

(√

)

1.175 kg/m3 =1.454 996 kg/m3

13

8. Calculate flow parameter, FLV (2 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.0779 2 m .s

(√ )

6.614

(√

)

1.175 kg /m3 =2.912 996 kg /m3

9. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.0779 2 m .s

(√ )

9.907

(√

)

1.175 kg /m3 =4.362 996 kg / m3

Theoretical Flooding Point for 40 L/min 3

40 L 1m 60 min 3 Gy = min × 1000 L × 1h =2.4 m / h Calculate gas flow rate, GG (kg/m2s)

1.

2.4 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.156 2 −3 2 5.027 ×10 m m .s

2.

Calculate capacity parameter, K4,

13.1 K4=

(

2

(

2 0.1

0.156 kg ( 0.001 N . s / m 900 m−1 ) 2 3 m s 996 kg /m

)

)

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

=0.062

3. Calculate flow parameter, FLV (1 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

3.302

(√

)

1.175 kg /m3 =0.727 996 kg/m3

14

4. Calculate flow parameter, FLV (2 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

6.614

(√

)

1.175 kg / m3 =1.456 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

9.907

(√

)

1.175 kg / m3 =2.181 996 kg/m 3

Theoretical Flooding Point for 60 L/min

60 L 1 m3 60 min 3 × Gy = min 1000 L × 1 h =3.6 m /h 1.

Calculate gas flow rate, GG (kg/m2s)

3.6 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.234 2 −3 2 5.027 ×10 m m .s

2.

Calculate capacity parameter, K4,

13.1 K4=

(

2 0.234 kg 2 ( −1 0.001 N . s / m ) 900 m m2 s 996 kg/m 3

)

(

0.1

)

( 1.175 kg /m3 )( 996 kg /m3−1.175 kg /m3 ) 15

=0.139

3. Calculate flow parameter, FLV (1 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

3.302

(√

)

1.175 kg / m3 =0.484 996 kg/m 3

4. Calculate flow parameter, FLV (2 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

6.614

(√

)

1.175 kg / m3 =0.969 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg m2 . s kg 0.156 2 m .s 9.907

(√

)

1.175 kg / m3 =1.452 3 996 kg/m

Theoretical Flooding Point for 80 L/min

80 L 1 m3 60 min 3 × Gy = min 1000 L × 1 h =4.8 m /h 1.

Calculate gas flow rate, GG (kg/m2s)

16

3

4.8 m 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.311 2 −3 2 5.027 ×10 m m .s

2.

Calculate capacity parameter, K4,

13.1 K4=

(

2 0.156 kg 2 ( −1 0.001 N . s /m ) 900 m m2 s 996 kg /m3

)

(

0.1

)

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

=0.245

3. Calculate flow parameter, FLV (1 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.311 2 m .s

(√ )

3.302

(√

)

1.175 kg/m3 =0.364 996 kg /m3

4. Calculate flow parameter, FLV (2 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.311 2 m .s

(√ )

6.614

(√

)

1.175 kg / m3 =0.729 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.311 2 m .s

(√ )

9.907

(√

)

1.175 kg /m3 =1.093 996 kg/m3

17

Theoretical Flooding Point for 100 L/min 3

100 L 1m 60 min 3 Gy = min × 1000 L × 1 h =6.0 m /h Calculate gas flow rate, GG (kg/m2s)

1.

6.0 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.389 2 −3 2 5.027 ×10 m m .s

2.

K4=

Calculate capacity parameter, K4, 2 0.156 kg 2 ( −1 0.001 N . s /m ) 13.1 900 m m2 s 996 kg /m3

(

)

(

0.1

)

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

=0.383

3. Calculate flow parameter, FLV (1 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg m2 . s kg 0.156 2 m .s 3.302

(√

)

1.175 kg / m3 =0.291 996 kg/m 3

4. Calculate flow parameter, FLV (2 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg m2 . s kg 0.156 2 m .s 6.614

(√

)

1.175 kg / m3 =0.583 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

9.907

(√

)

1.175 kg / m3 =0.874 3 996 kg/m

18

Theoretical Flooding Point for 120 L/min

120 L 1 m3 60 min 3 × Gy = min 1000 L × 1 h =7.2 m /h 1.

Calculate gas flow rate, GG (kg/m2s) 3

7.2m 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.467 2 −3 2 5.027 ×10 m m .s

2.

K4=

Calculate capacity parameter, K4, 2 0.467 kg 2 ( −1 0.001 N . s / m ) 13.1 900 m m2 s 996 kg /m 3

(

)

(

0.1

)

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

=0.553

3. Calculate flow parameter, FLV (1 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg 2 m .s kg 0.467 2 m .s 3.302

(√

3

)

1.175 kg /m =0.243 996 kg/m3

4. Calculate flow parameter, FLV (2 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg m2 . s kg 0.467 2 m .s 6.614

(√

)

1.175 kg / m3 =0.486 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

19

FLV =

GL GG

kg 2 ρG m .s = ρL kg 0.467 2 m .s

(√ )

9.907

(√

3

)

1.175 kg / m =0.728 996 kg/m 3

Theoretical Flooding Point for 140 L/min

140 L 1 m3 60 min 3 × Gy = min 1000 L × 1 h =8.4 m /h Calculate gas flow rate, GG (kg/m2s)

1.

3

8.4 m 1h 1.175 kg × × h 3600 s m3 kg G G= =0.545 2 −3 2 5.027 ×10 m m .s

2.

K4=

Calculate capacity parameter, K4, 2

(

0.545 kg ( 0.001 N . s /m 13.1 900 m−1 ) 2 3 m s 996 kg /m

(

)

2 0.1

)

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

=0.753

3. Calculate flow parameter, FLV (1 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

3.302

(√

)

1.175 kg /m3 =0.208 996 kg/m3

4. Calculate flow parameter, FLV (2 LPM) 20

FLV =

GL GG

kg 2 ρG m .s = ρL kg 0.156 2 m .s

(√ )

6.614

(√

3

)

1.175 kg / m =0.416 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.156 2 m .s

(√ )

9.907

(√

)

1.175 kg / m3 =0.624 996 kg/m 3

Theoretical Flooding Point for 160 L/min

160 L 1 m3 60 min 3 Gy = min × 1000 L × 1 h =9.6 m /h 1.

Calculate gas flow rate, GG (kg/m2s)

9.6 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.623 2 −3 2 5.027 ×10 m m .s

2.

K4=

Calculate capacity parameter, K4, 2

(

0.623 kg ( 0.001 N . s /m 13.1 900 m−1 ) 2 3 m s 996 kg /m

(

)

2 0.1

)

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

=0.989

3. Calculate flow parameter, FLV (1 LPM) 21

FLV =

GL GG

kg 2 ρG m .s = ρL kg 0.623 2 m .s

(√ )

3.302

(√

3

)

1.175 kg /m =0.182 996 kg /m3

4. Calculate flow parameter, FLV (2 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.623 2 m .s

(√ )

6.614

(√

)

1.175 kg / m3 =0.364 996 kg/m 3

5. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg ρG m2 . s = ρL kg 0.623 2 m .s

(√ )

9.907

(√

)

1.175 kg / m3 =0.545 996 kg/m 3

Theoretical Flooding Point for 1800 L/min 3

180 L 1m 60 min 3 Gy = min × 1000 L × 1 h =10.8 m /h 1.

Calculate gas flow rate, GG (kg/m2s)

10.8 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.701 2 −3 2 5.027 ×10 m m .s

22

Calculate capacity parameter, K4,

2.

K4=

2 0.701 kg 2 ( −1 0.001 N . s/m ) 13.1 900m m2 s 996 kg /m3

(

)

(

0.1

)

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

=1.245

3. Calculate flow parameter, FLV (1 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg 2 m .s kg 0.701 2 m .s 3.302

(√

3

)

1.175 kg /m =0.162 3 996 kg /m

4. Calculate flow parameter, FLV (2 LPM)

G FLV = L GG

(√ )

ρG = ρL

kg 2 m .s kg 0.701 2 m .s 6.614

(√

3

)

1.175 kg / m =0.324 3 996 kg/m

5. Calculate flow parameter, FLV (3 LPM)

FLV =

GL GG

kg 2 ρG m .s = ρL kg 0.701 2 m .s

(√ )

9.907

(√

3

)

1.175 kg / m =0.485 3 996 kg/m

PERCENTAGE ERROR % 1LPM

23

Total correlated pressure drop = 243.38 mm H20 Total pressure drop = 126 mm H20

Percentage error , =

¿

theoritical value−experimental value × 100 theoritical value

244.1−126 ×100 =48 244.1

2LPM Total correlated pressure drop = 177.8 mm H20 Total pressure drop = 105 mm H20

Percentage error , =

¿

theoritical value−experimental value × 100 theoritical value

233.6−105 × 100 =46 233.6

3LPM Total correlated pressure drop = 180.3mm H20 Total pressure drop =61 mm H20

Percentage error , =

¿

theoritical value−experimental value × 100 theoritical value

180.34−61 × 100 =66 180.34

24

DISCUSSION For this experiment, the aim is to determine the pressure drop correlation for different flow rate of water 1 LPM, 2 LPM and 3 LPM. The air flow rate increased as the pressure drop in the dry packed column increases. These occur due to the air flow rate increased results of increasing in resistance for the water to flows down the column and give high pressure drop across the packing. In the graph of log pressure drop against log gas flow rate, the log air flow rate increase as the log pressure drop is also increase. The air flow rate increase as the pressure drop increase in constant flow of water. At 1 LPM, the pressure drop is the lowest water flow rate compared to the other two flow rate. This is because of the space for gas flow is blocked by the liquid that flows sown the column. The water flows down due to the gravitational force and thus the gas flows in a counter-current direction with water. During this experiment, there are some resistances occur in the column such as the gas from the bottom starts to impede the water flowing down the column and overflow of water. Because of that, the graph in figure 1 differ from theoretically graph which is 1 LPM should be a straight line while the other two flow rate which is 2 LPM and 3 LPM should be parallel to the graph 1 LPM. From the calculation of K 4 and FLV, the plotted graph is obtained. The relationship between the plotted graphs is K4 is inversely proportional to FLV. As the value of FLV increase, the value of K4 decreases. This shows that the relationship of pressure drop and gas flow rate is increasing linearly. Then those values can use to generalize correlation for pressure drop in packed column in chart shows in Appendix. Theoretical generalized correlation charts show that the high flow parameters are typical of high liquid rates and high pressure drop. However, by looking at both of graph, it shows difference of value pressure drop in theoretical and experimental. 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 point is the highest point for each line in the graph of pressure against gas flow rate. When this happen, the process can be no longer be conducted because there is too 25

much liquid entrainment. The flooding points occur at 180 L/min,140 L/min in and 100 L/min for 1LPM,2LPM and 3LPM respectively. CONCLUSION In conclusion, the pressure drop will increase when the gas flow increased at constant water flow rate of 1 LPM to 3 LPM. The gas was absorbed through the packed column in a batch process in absorption of air and the effect of liquid flow rate on the absorption-adsorption process was observed. As the flow rate of air increased, the absorptionadsorption process also increased as the composition of the outlet volume of air increased over time. The resistance to of water flows down the column due to the increasing water flow rate. As for the theoretical value, the same principle is used based on the pressure drop correlation charts as well as the experimental value. However, there are are different value in pressure drops value as there are some error occur which can prevent in recommendation. From the experiment, the value of experimental pressure drop is lower compared to the correlated values for packed column.

RECOMMENDATION There are some recommendations that should be taken account into to ensure the experiment to become more accurate which are the valve controlling the level of water flowing back to the water reservoir should be constantly checked so that we can get a better reading. The level of water must be higher than the bottom of the reservoir. This need to be done to avoid air being trapped in line. Beside that, Make sure all the valves are closed before using the column so that the experiment runs smoothly. Moreover, Make sure the gas and liquid flow rates were constant at that particular flow rate Then, the gas and liquid flow rates must be constant at that particular flow rates. Then, collect the samples simultaneously from both inlet and outlet of the packed column. Furthermore, Give the experiment some more time before the results are taken.

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REFERENCES 1. Transport Process and Separation Process Principles (Includes Unit Operations) 4th Edition, Christie John Geankoplis, Pearson Education Inc 2. Principle of Gas Absorption retrieved from http://pubs.acs.org/doi/abs/10.1021/ie50180a002 3. Gas Absorption Lab Manual 4. http://separationprocess.com/ 5. Principle of Gas Absorption retrieved from http://pubs.acs.org/doi/abs/10.1021/ie50180a002

APPENDIX

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