e533 Gas Absorption Full Lab Report 2vn4t
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UNIVERSITI TEKNOLOGI MARA FAKULTY OF CHEMICAL ENGINEERING PROCESS ENGINEERING LABORATORY 2 (E553) NAME : MOHAMAD FAZRUL BIN BASRI (2013331897) GROUP : EH2414 (GROUP1) EXPERIM : LAB 11 : GAS ABSORPTION ENT DATE : 6 NOVEMBER 2015 PROG/CO : EH241 DE SUBMIT : MDM LIM YING PEI TO N Title Allocated Marks o Marks (%) 1 Abstract 5 2 Introduction 5 3 Objectives 5 4 Theory 5 5 Procedures/Methodology 10 6 Apparatus 5 7 Results 10 8 Calculation 10 9 Discussion 20 10 Conclusion 10 11 Recommendations 5 12 References 5 13 Appendices 5 TOTAL 100 Remarks:
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ABSTRACT The Gas Absorption experiment was conducted in order to examine the air pressure drop across the absorption column as a function of air flow rate with different rates of water flow. Gas absorption is a process in which a gaseous mixture is brought into with a liquid and during this a component is transferred between the gas stream and the liquid stream. The gas may be bubbled through the liquid, or it may 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. 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 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 LPM until 3 LPM. The gas flow rate starts from 20 L/min until 180 L/min with 10 L/min of intervals. How fast the liquid can flow down with no vapour flowing upwards and the rate at which the vapour is trying to flow upwards is the actual flooding point. The flooding point was recorded during the water flow rate of 1 LPM, 2 LPM and 3 LPM which were 160 L/min, 120 L/min and 80 L/min respectively.
INTRODUCTION 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. An example of gas is the absorption of the solute ammonia from an air-ammonia mixture by water. A major application of absorption is the removal of CO2 and H2S from nature gas or synthesis gas by absorption in solution of amines or alkaline salts. A common apparatus used in gas absorption and certain other operations is the packed tower, shown in Figure 1 below. 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 ed mass of inert solid shapes, called tower packing.
Figure 1: Gas Absorption Packing Column
Their common dumped packing, Ceramic Berl saddles and Raschig rings are older types of packing that are not much used now, although there were big improvements over ceramic spheres or crushed stone when first introduced. The shape prevents pieces from nesting closely together, and this increasing the bed porosity. As for this experiment we used the column packed with Raschig rings. 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 or flooding point. Above this gas velocity the tower cannot operate due to high pressure. 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.
OBJECTIVES To examine the air pressure drop across the column as a function of air flow for different water flow rates through the column. 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 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 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, 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)
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 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 around 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 was turned off.
RESULTS Pressure Drop
Flow rate
(mm
H2O ¿
(L/min) air water
20
40
1.0 2.0 3.0
0 25 27
1 27 31
60
80
100
120
140
160
180
3 15 26 28 32 31 35 41 51 36 47 Table 1: Pressure Drop for Wet column
44 -
-
Flow rate Air
Gas Flow rate 20 40 60 80 100 120 140 160 180
Water
Log Gas Flow rate 1.301 1.602 1.778 1.903 2.00 2.079 2.146 2.204 2.255
1.0
Pressure drop (mmH2O)
2.0
Log Pressure drop
Pressure drop (mmH2O)
3.0
Log Pressure drop
Pressure drop (mmH2O)
(mmH2O) (mmH2O) 0 25 1.398 1 0.000 27 1.431 3 0.477 31 1.491 15 1.176 35 1.544 26 1.415 41 1.613 28 1.447 51 1.708 32 1.505 44 1.643 Table 2: Log Gas Flow rate and Log Pressure drop
27 31 36 47 -
Log Pressure drop (mmH2O) 1.431 1.491 1.556 1.672 -
Air
Air
Flow
Flow
GG 2
(kg/ms )
Pressure drop
FLV
FLV
FLV
(1
(2
(3
LPM)
LPM)
LPM)
1LPM
2LPM
3LPM
1.454 0.727
2.912
4.362
1.52
5.08
12.7
1.456
2.181
5.08
10.16
25.4
0.139 0.484 0.969 1.452 0.245 0.364 0.729 1.093 0.383 0.291 0.583 0.553 0.243 0.486 0.753 0.208 0.984 0.182 1.245 Table 3: Theoretical Flooding Point
8.89 12.7 25.4 40.64 43.18 50.80 -
25.4 40.64 45.72 50.8 -
38.1 50.8 -
K4
rate
rate
(L/min)
(m3/h)
20
1.2
0.0779
0.0154
40
2.4
0.156
0.062
60 80 100 120 140 160 180
3.6 4.8 6.0 7.2 8.4 9.6 10.8
0.234 0.311 0.389 0.467 0.545 0.623 0.701
Log Pressure drop correlated in mm H20 1LPM
2LPM
3LPM
0.18
0.71
1.10
0.71
1.00
1.40
0.95
1.40
1.58
1.10
1.61
1.71
1.40
1.66
-
1.61
1.71
-
1.64
-
-
1.71
-
-
Table 4: Log Pressure drop correlated in mm H20
correlated in mm H20
Graph of Log Pressure Drop against Log Gas Flow Rate
Log Pressure Drop, mmH2O
5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0
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
log correlated pressure drop
2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0
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
1.2 m3 1h 1.175 kg × × h 3600 s m3 kg ¿ =0.0779 2 −3 2 5.027 ×10 m m .s
3.
Calculate capacity parameter, K4,
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) 3
1 L 1 min 1m 996 kg × × × 3 G× ρ min 60 s 1000l m GL = = −3 2 Ac 5.027 ×10 m
¿ 3.302
kg 2 m .s
5. Calculate liquid flow rate, GL (kg/m2) (2 LPM)
2 L 1 mi n 1 m3 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 L 1 min 1m3 996 kg × × × min 60 s 1000l G× ρ m3 kg GL = = =9.907 2 −3 2 Ac 5.027 ×10 m m .s 7. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.0779 2 m .s
(√ )
3.302
(√
3
)
3
)
1.175 kg /m =1.454 996 kg / m3
8. Calculate flow parameter, FLV (2 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.0779 2 m .s
(√ )
6.614
(√
1.175 kg /m =2.912 996 kg /m3
9. Calculate flow parameter, FLV (3 LPM)
G FLV = L GG
(√ )
ρG = ρL
kg m2 . s kg 0.0779 2 m .s 9.907
(√
Theoretical Flooding Point for 40 L/min
40 L 1 m3 60 min 3 × Gy = min 1000 L × 1h =2.4 m /h 1.
Calculate gas flow rate, GG (kg/m2s)
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.
)
1.175 kg /m3 =4.362 996 kg / m3
Calculate capacity parameter, K4,
K4=
2
2 0.1
(
0.156 kg ( 0.001 N . s / m 13.1 900 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
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 =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
Calculate gas flow rate, GG (kg/m2s)
1.
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.
K4=
Calculate capacity parameter, K4, 2 0.234 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 )
=0.139
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.156 2 m .s
(√ )
3.302
(√
3
)
1.175 kg / m =0.484 3 996 kg/m
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 =1.452 996 kg/m 3
Theoretical Flooding Point for 80 L/min
80 L 1 m3 60 min 3 × Gy = min 1000 L × 1 h =4.8 m /h Calculate gas flow rate, GG (kg/m2s)
1.
4.8 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.311 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)
=0.245
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)
G FLV = L GG
(√ )
ρG = ρL
kg m2 . s 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 2 ρG m .s = ρL kg 0.311 2 m .s
(√ )
9.907
(√
3
)
1.175 kg /m =1.093 996 kg/m3
Theoretical Flooding Point for 100 L/min
100 L 1 m3 60 min × × =6.0 m3 /h Gy = min 1000 L 1h 1.
Calculate gas flow rate, GG (kg/m2s) 3
6.0 m 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.389 2 −3 2 5.027 ×10 m m .s
2.
K4=
Calculate capacity parameter, K4, 2
(
2 0.1
0.156 kg ( 0.001 N . s /m 13.1 900 m−1 ) 2 3 m s 996 kg /m
(
)
)
( 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)
FLV =
GL GG
kg ρG m2 . s = ρL 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)
G FLV = L GG
(√ )
ρG = ρL
kg 2 m .s kg 0.156 2 m .s 9.907
(√
3
Theoretical Flooding Point for 120 L/min
120 L 1 m3 60 min × × =7.2 m3 /h Gy = min 1000 L 1h 1.
Calculate gas flow rate, GG (kg/m2s)
7.2m3 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.467 2 −3 2 5.027 ×10 m m .s
2.
)
1.175 kg / m =0.874 996 kg/m 3
Calculate capacity parameter, K4,
K4=
2
2 0.1
(
0.467 kg ( 0.001 N . s / m 13.1 900 m−1 ) 2 3 m s 996 kg /m
(
)
)
( 1.175 kg /m3 ) ( 996 kg/m3−1.175 kg /m3)
=0.553
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.467 2 m .s
(√ )
3.302
(√
)
1.175 kg /m3 =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)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.467 2 m .s
(√ )
9.907
(√
)
1.175 kg / m3 =0.728 996 kg/m 3
Theoretical Flooding Point for 140 L/min
140 L 1 m3 60 min × × =8.4 m3 /h Gy = min 1000 L 1h
Calculate gas flow rate, GG (kg/m2s)
1.
8.4 m3 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 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 )
=0.753
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.156 2 m .s
(√ )
3.302
(√
3
)
1.175 kg /m =0.208 3 996 kg/m
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 =0.416 996 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 =0.624 996 kg/m 3
Theoretical Flooding Point for 160 L/min
160 L 1 m3 60 min × × =9.6 m3 /h Gy = min 1000 L 1h Calculate gas flow rate, GG (kg/m2s)
1.
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 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 )
=0.989
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.623 2 m .s
(√ )
3.302
(√
)
1.175 kg /m3 =0.182 996 kg /m3
4. Calculate flow parameter, FLV (2 LPM)
G FLV = L GG
(√ )
ρG = ρL
kg m2 . s 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 2 ρG m .s = ρL kg 0.623 2 m .s
(√ )
9.907
(√
)
3
1.175 kg / m =0.545 996 kg/m 3
Theoretical Flooding Point for 180 L/min
180 L 1 m3 60 min × × =10.8 m3 /h Gy = min 1000 L 1h 1.
Calculate gas flow rate, GG (kg/m2s) 3
10.8 m 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.701 2 −3 2 5.027 ×10 m m .s
2.
K4=
Calculate capacity parameter, K4, 2
(
0.701 kg ( 0.001 N . s/m 13.1 900m−1 ) 2 3 m s 996 kg /m
(
)
2 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 m2 . s kg 0.701 2 m .s 3.302
(√
)
1.175 kg /m3 =0.162 996 kg /m3
4. Calculate flow parameter, FLV (2 LPM)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.701 2 m .s
(√ )
6.614
(√
)
1.175 kg / m3 =0.324 996 kg/m 3
5. Calculate flow parameter, FLV (3 LPM)
G FLV = L GG
(√ )
ρG = ρL
kg 2 m .s kg 0.701 2 m .s 9.907
(√
3
)
1.175 kg / m =0.485 996 kg/m 3
Percentage Error % 1LPM Total correlated pressure drop = 185.21 mm H20 Total pressure drop = 149.00 mm H20
Percentage error , =
¿∨
theoritical value−experimental value × 100 theoritical value
185.21−149.00 ∨× 100 =19.55 185.21
2LPM Total correlated pressure drop = 177.80 mm H20 Total pressure drop = 210.00 mm H20
Percentage error , =
¿∨
theoritical value−experimental value × 100 theoritical value
177.80−210.00 ∨×100 =18.08 177.80
3LPM Total correlated pressure drop = 127.00 mm H20 Total pressure drop =141.00 mm H20
Percentage error , =
¿∨
theoritical value−experimental value × 100 theoritical value
127.00−141.00 ∨×100 =11.02 127.00
DISCUSSION The important things that we want to find out from the experiment are to determine where is the flooding and loading point of the gas absorption as well as to determine the pressure drop (∆P) as a function of gas (air) and liquid (water) mass velocities (m 3/h) using flexi glass packed with Raschig rings. 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. The following are the graphs showing the pattern on the pressure drop with different value of water flow rate; 1 LPM, 2 LPM and 3. LPM. During the experiment, 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 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 19.55%, 2LPM is 18.08% and 3LPM is 11.02%. It 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 much liquid entrainment. The flooding points occur at 160 L/min, 120 L/min and 80 L/min for 1 LPM, 2 LPM and 3 LPM respectively.
CONCLUSION At the end of the experiment, we had managed to determine the loading and flooding point where the loading point is from the volume water flow rate of 1 LPM, 2 LPM and 3 LPM. 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. When the volume water flow rate at 1 LPM, 2 LPM and 3 LPM, the flooding point started at gas flow rate of 160 L/min, 120 L/min and 80 L/min respectively. As the flow rate of air increased, the absorption-adsorption process also increased as the composition of the outlet volume of air increased over time. Thus, we can conclude that all the objectives of the experiment had been reached. RECOMMENDATIONS During the experiments, there are some mistakes that have been made. Those mistakes had affected the result of experiment which is the pressure drop (∆P)as a function of gas (air) and liquid (water)mass velocities ((L/min)) using flexi glass packed with Raschig rings. The mistakes are caused by parallax error when reading the pressure on the monotube and also due to the lack of control on the water pump. Apart from that, to ensure the experiment to become more accurate, 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 needs to be done to avoid the air being trapped in line. Besides 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.
REFERENCES
C.J. Geankoplis.Transport Processes and Separation Process Principles(includes unit
operations). Fourth edition. Page 119 CAG Gas Absorption Column.
Retrieved
June
1,
2015
from
http://www.edibon.com/products/catalogues/en/units/chemicalengineering/chemicalengi
neeringbasic/CAG.pdf Chemical Engineering:
http://chem.engr.utc.edu/Webres/435F/ABS_COL/abs_col.html Dr. Rami Jumah (2002). Unit Operation Laboratory. Jordan University of Science and
Technology. Retrieved on 20th April 2013 Gas Absorption handouts. Retrieved
http://www.engr.uconn.edu/~ewanders/CHEG237W/Gas-Absorption.pdf Richardson, J. F. and Harker, J. H. (2002). Chemical Engineering. Fifth Edition. Page
655. Yunus A. Cengel et. al., Fluid Mechanics Fundamentals and Applications, 2 nd Edition, McGraw Hill.
APPENDICES
Gas
Absorption
Column.
June
June
1,
1,
2015
2015
from
from
Checked by:
Rechecked by:
Date:
Date:
ABSTRACT The Gas Absorption experiment was conducted in order to examine the air pressure drop across the absorption column as a function of air flow rate with different rates of water flow. Gas absorption is a process in which a gaseous mixture is brought into with a liquid and during this a component is transferred between the gas stream and the liquid stream. The gas may be bubbled through the liquid, or it may 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. 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 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 LPM until 3 LPM. The gas flow rate starts from 20 L/min until 180 L/min with 10 L/min of intervals. How fast the liquid can flow down with no vapour flowing upwards and the rate at which the vapour is trying to flow upwards is the actual flooding point. The flooding point was recorded during the water flow rate of 1 LPM, 2 LPM and 3 LPM which were 160 L/min, 120 L/min and 80 L/min respectively.
INTRODUCTION 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. An example of gas is the absorption of the solute ammonia from an air-ammonia mixture by water. A major application of absorption is the removal of CO2 and H2S from nature gas or synthesis gas by absorption in solution of amines or alkaline salts. A common apparatus used in gas absorption and certain other operations is the packed tower, shown in Figure 1 below. 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 ed mass of inert solid shapes, called tower packing.
Figure 1: Gas Absorption Packing Column
Their common dumped packing, Ceramic Berl saddles and Raschig rings are older types of packing that are not much used now, although there were big improvements over ceramic spheres or crushed stone when first introduced. The shape prevents pieces from nesting closely together, and this increasing the bed porosity. As for this experiment we used the column packed with Raschig rings. 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 or flooding point. Above this gas velocity the tower cannot operate due to high pressure. 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.
OBJECTIVES To examine the air pressure drop across the column as a function of air flow for different water flow rates through the column. 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 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 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, 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)
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 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 around 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 was turned off.
RESULTS Pressure Drop
Flow rate
(mm
H2O ¿
(L/min) air water
20
40
1.0 2.0 3.0
0 25 27
1 27 31
60
80
100
120
140
160
180
3 15 26 28 32 31 35 41 51 36 47 Table 1: Pressure Drop for Wet column
44 -
-
Flow rate Air
Gas Flow rate 20 40 60 80 100 120 140 160 180
Water
Log Gas Flow rate 1.301 1.602 1.778 1.903 2.00 2.079 2.146 2.204 2.255
1.0
Pressure drop (mmH2O)
2.0
Log Pressure drop
Pressure drop (mmH2O)
3.0
Log Pressure drop
Pressure drop (mmH2O)
(mmH2O) (mmH2O) 0 25 1.398 1 0.000 27 1.431 3 0.477 31 1.491 15 1.176 35 1.544 26 1.415 41 1.613 28 1.447 51 1.708 32 1.505 44 1.643 Table 2: Log Gas Flow rate and Log Pressure drop
27 31 36 47 -
Log Pressure drop (mmH2O) 1.431 1.491 1.556 1.672 -
Air
Air
Flow
Flow
GG 2
(kg/ms )
Pressure drop
FLV
FLV
FLV
(1
(2
(3
LPM)
LPM)
LPM)
1LPM
2LPM
3LPM
1.454 0.727
2.912
4.362
1.52
5.08
12.7
1.456
2.181
5.08
10.16
25.4
0.139 0.484 0.969 1.452 0.245 0.364 0.729 1.093 0.383 0.291 0.583 0.553 0.243 0.486 0.753 0.208 0.984 0.182 1.245 Table 3: Theoretical Flooding Point
8.89 12.7 25.4 40.64 43.18 50.80 -
25.4 40.64 45.72 50.8 -
38.1 50.8 -
K4
rate
rate
(L/min)
(m3/h)
20
1.2
0.0779
0.0154
40
2.4
0.156
0.062
60 80 100 120 140 160 180
3.6 4.8 6.0 7.2 8.4 9.6 10.8
0.234 0.311 0.389 0.467 0.545 0.623 0.701
Log Pressure drop correlated in mm H20 1LPM
2LPM
3LPM
0.18
0.71
1.10
0.71
1.00
1.40
0.95
1.40
1.58
1.10
1.61
1.71
1.40
1.66
-
1.61
1.71
-
1.64
-
-
1.71
-
-
Table 4: Log Pressure drop correlated in mm H20
correlated in mm H20
Graph of Log Pressure Drop against Log Gas Flow Rate
Log Pressure Drop, mmH2O
5 4.5 4 3.5 3 2.5 2 1.5 1 0.5 0
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
log correlated pressure drop
2 1.8 1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0
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
1.2 m3 1h 1.175 kg × × h 3600 s m3 kg ¿ =0.0779 2 −3 2 5.027 ×10 m m .s
3.
Calculate capacity parameter, K4,
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) 3
1 L 1 min 1m 996 kg × × × 3 G× ρ min 60 s 1000l m GL = = −3 2 Ac 5.027 ×10 m
¿ 3.302
kg 2 m .s
5. Calculate liquid flow rate, GL (kg/m2) (2 LPM)
2 L 1 mi n 1 m3 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 L 1 min 1m3 996 kg × × × min 60 s 1000l G× ρ m3 kg GL = = =9.907 2 −3 2 Ac 5.027 ×10 m m .s 7. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.0779 2 m .s
(√ )
3.302
(√
3
)
3
)
1.175 kg /m =1.454 996 kg / m3
8. Calculate flow parameter, FLV (2 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.0779 2 m .s
(√ )
6.614
(√
1.175 kg /m =2.912 996 kg /m3
9. Calculate flow parameter, FLV (3 LPM)
G FLV = L GG
(√ )
ρG = ρL
kg m2 . s kg 0.0779 2 m .s 9.907
(√
Theoretical Flooding Point for 40 L/min
40 L 1 m3 60 min 3 × Gy = min 1000 L × 1h =2.4 m /h 1.
Calculate gas flow rate, GG (kg/m2s)
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.
)
1.175 kg /m3 =4.362 996 kg / m3
Calculate capacity parameter, K4,
K4=
2
2 0.1
(
0.156 kg ( 0.001 N . s / m 13.1 900 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
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 =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
Calculate gas flow rate, GG (kg/m2s)
1.
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.
K4=
Calculate capacity parameter, K4, 2 0.234 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 )
=0.139
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.156 2 m .s
(√ )
3.302
(√
3
)
1.175 kg / m =0.484 3 996 kg/m
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 =1.452 996 kg/m 3
Theoretical Flooding Point for 80 L/min
80 L 1 m3 60 min 3 × Gy = min 1000 L × 1 h =4.8 m /h Calculate gas flow rate, GG (kg/m2s)
1.
4.8 m3 1h 1.175 kg × × h 3600 s m3 kg G G= =0.311 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)
=0.245
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)
G FLV = L GG
(√ )
ρG = ρL
kg m2 . s 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 2 ρG m .s = ρL kg 0.311 2 m .s
(√ )
9.907
(√
3
)
1.175 kg /m =1.093 996 kg/m3
Theoretical Flooding Point for 100 L/min
100 L 1 m3 60 min × × =6.0 m3 /h Gy = min 1000 L 1h 1.
Calculate gas flow rate, GG (kg/m2s) 3
6.0 m 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.389 2 −3 2 5.027 ×10 m m .s
2.
K4=
Calculate capacity parameter, K4, 2
(
2 0.1
0.156 kg ( 0.001 N . s /m 13.1 900 m−1 ) 2 3 m s 996 kg /m
(
)
)
( 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)
FLV =
GL GG
kg ρG m2 . s = ρL 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)
G FLV = L GG
(√ )
ρG = ρL
kg 2 m .s kg 0.156 2 m .s 9.907
(√
3
Theoretical Flooding Point for 120 L/min
120 L 1 m3 60 min × × =7.2 m3 /h Gy = min 1000 L 1h 1.
Calculate gas flow rate, GG (kg/m2s)
7.2m3 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.467 2 −3 2 5.027 ×10 m m .s
2.
)
1.175 kg / m =0.874 996 kg/m 3
Calculate capacity parameter, K4,
K4=
2
2 0.1
(
0.467 kg ( 0.001 N . s / m 13.1 900 m−1 ) 2 3 m s 996 kg /m
(
)
)
( 1.175 kg /m3 ) ( 996 kg/m3−1.175 kg /m3)
=0.553
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.467 2 m .s
(√ )
3.302
(√
)
1.175 kg /m3 =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)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.467 2 m .s
(√ )
9.907
(√
)
1.175 kg / m3 =0.728 996 kg/m 3
Theoretical Flooding Point for 140 L/min
140 L 1 m3 60 min × × =8.4 m3 /h Gy = min 1000 L 1h
Calculate gas flow rate, GG (kg/m2s)
1.
8.4 m3 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 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 )
=0.753
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg 2 ρG m .s = ρL kg 0.156 2 m .s
(√ )
3.302
(√
3
)
1.175 kg /m =0.208 3 996 kg/m
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 =0.416 996 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 =0.624 996 kg/m 3
Theoretical Flooding Point for 160 L/min
160 L 1 m3 60 min × × =9.6 m3 /h Gy = min 1000 L 1h Calculate gas flow rate, GG (kg/m2s)
1.
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 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 )
=0.989
3. Calculate flow parameter, FLV (1 LPM)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.623 2 m .s
(√ )
3.302
(√
)
1.175 kg /m3 =0.182 996 kg /m3
4. Calculate flow parameter, FLV (2 LPM)
G FLV = L GG
(√ )
ρG = ρL
kg m2 . s 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 2 ρG m .s = ρL kg 0.623 2 m .s
(√ )
9.907
(√
)
3
1.175 kg / m =0.545 996 kg/m 3
Theoretical Flooding Point for 180 L/min
180 L 1 m3 60 min × × =10.8 m3 /h Gy = min 1000 L 1h 1.
Calculate gas flow rate, GG (kg/m2s) 3
10.8 m 1h 1.175 kg × × 3 h 3600 s m kg G G= =0.701 2 −3 2 5.027 ×10 m m .s
2.
K4=
Calculate capacity parameter, K4, 2
(
0.701 kg ( 0.001 N . s/m 13.1 900m−1 ) 2 3 m s 996 kg /m
(
)
2 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 m2 . s kg 0.701 2 m .s 3.302
(√
)
1.175 kg /m3 =0.162 996 kg /m3
4. Calculate flow parameter, FLV (2 LPM)
FLV =
GL GG
kg ρG m2 . s = ρL kg 0.701 2 m .s
(√ )
6.614
(√
)
1.175 kg / m3 =0.324 996 kg/m 3
5. Calculate flow parameter, FLV (3 LPM)
G FLV = L GG
(√ )
ρG = ρL
kg 2 m .s kg 0.701 2 m .s 9.907
(√
3
)
1.175 kg / m =0.485 996 kg/m 3
Percentage Error % 1LPM Total correlated pressure drop = 185.21 mm H20 Total pressure drop = 149.00 mm H20
Percentage error , =
¿∨
theoritical value−experimental value × 100 theoritical value
185.21−149.00 ∨× 100 =19.55 185.21
2LPM Total correlated pressure drop = 177.80 mm H20 Total pressure drop = 210.00 mm H20
Percentage error , =
¿∨
theoritical value−experimental value × 100 theoritical value
177.80−210.00 ∨×100 =18.08 177.80
3LPM Total correlated pressure drop = 127.00 mm H20 Total pressure drop =141.00 mm H20
Percentage error , =
¿∨
theoritical value−experimental value × 100 theoritical value
127.00−141.00 ∨×100 =11.02 127.00
DISCUSSION The important things that we want to find out from the experiment are to determine where is the flooding and loading point of the gas absorption as well as to determine the pressure drop (∆P) as a function of gas (air) and liquid (water) mass velocities (m 3/h) using flexi glass packed with Raschig rings. 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. The following are the graphs showing the pattern on the pressure drop with different value of water flow rate; 1 LPM, 2 LPM and 3. LPM. During the experiment, 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 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 19.55%, 2LPM is 18.08% and 3LPM is 11.02%. It 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 much liquid entrainment. The flooding points occur at 160 L/min, 120 L/min and 80 L/min for 1 LPM, 2 LPM and 3 LPM respectively.
CONCLUSION At the end of the experiment, we had managed to determine the loading and flooding point where the loading point is from the volume water flow rate of 1 LPM, 2 LPM and 3 LPM. 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. When the volume water flow rate at 1 LPM, 2 LPM and 3 LPM, the flooding point started at gas flow rate of 160 L/min, 120 L/min and 80 L/min respectively. As the flow rate of air increased, the absorption-adsorption process also increased as the composition of the outlet volume of air increased over time. Thus, we can conclude that all the objectives of the experiment had been reached. RECOMMENDATIONS During the experiments, there are some mistakes that have been made. Those mistakes had affected the result of experiment which is the pressure drop (∆P)as a function of gas (air) and liquid (water)mass velocities ((L/min)) using flexi glass packed with Raschig rings. The mistakes are caused by parallax error when reading the pressure on the monotube and also due to the lack of control on the water pump. Apart from that, to ensure the experiment to become more accurate, 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 needs to be done to avoid the air being trapped in line. Besides 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.
REFERENCES
C.J. Geankoplis.Transport Processes and Separation Process Principles(includes unit
operations). Fourth edition. Page 119 CAG Gas Absorption Column.
Retrieved
June
1,
2015
from
http://www.edibon.com/products/catalogues/en/units/chemicalengineering/chemicalengi
neeringbasic/CAG.pdf Chemical Engineering:
http://chem.engr.utc.edu/Webres/435F/ABS_COL/abs_col.html Dr. Rami Jumah (2002). Unit Operation Laboratory. Jordan University of Science and
Technology. Retrieved on 20th April 2013 Gas Absorption handouts. Retrieved
http://www.engr.uconn.edu/~ewanders/CHEG237W/Gas-Absorption.pdf Richardson, J. F. and Harker, J. H. (2002). Chemical Engineering. Fifth Edition. Page
655. Yunus A. Cengel et. al., Fluid Mechanics Fundamentals and Applications, 2 nd Edition, McGraw Hill.
APPENDICES
Gas
Absorption
Column.
June
June
1,
1,
2015
2015
from
from
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