Basra University / College Of Engineering
Chemical Engineering Department
fourth Stage
Unit Laboratory
Experiment : fluid flow through packed column
Date of experiment : 2015/3/7
Student :
Mohammed Kadhim Mohammed
NO : (64)
The purpose of experiment:
To find the difference between practical and theoretical calculation of the
pressure drop through packed bed column change with time .
INTRODUCTION:
The flow of fluids through packed columns is a frequent occurrence in the
chemical industry. Fixed bed catalyst reactors, packed bed adsorption
columns, drying columns packed with silica gel or molecular sieves-etc. are
a few examples of equipment where fluids have to pass through packed
columns. Therefore it is necessary to obtain expressions for the prediction
of the pressure drop across the packing materials . in this experiment air is
passed through a column packed with Rashing rings. The pressure drop
across two sections is determined by Ergun and Orning combined two
different models, one accounting for viscous energy losses, and another
accounting for kinetic energy losses, to create the Ergun equation.
Pressure drop:
It is important to be able to predict the drop in pressure for the flow of the
two fluid streams through a packed column. Earlier in this chapter the drop
in pressure arising from the flow of a single phase through granular beds is
considered and the same general form of approach is usefully adopted for
the flow of two fluids through packed columns. It was noted that the
expressions for flow through ring-type packing are less reliable than those
for flow through beds of solid particles. For the typical absorption column
there is no very accurate expression, but there are several correlations that
are useful for design purposes. In the majority of cases the gas flow is
turbulent and the general form of the relation between the drop in pressure
P and the volumetric gas flow rate per unit area of column .
Types of liquid distributor:
Theory:
The work of Carman and Kozeny, as well as Blake and Plummer, formed
the basis for Ergun’s initial experiments on gas flow through packed beds
of crushed, porous solids at Carnegie:
∆P, K1, L, Ԑ, μ, U, gc and DP are the change in pressure, coefficient of
viscous energy, total height of the packed bed, void fraction, fluid viscosity,
superficial velocity, a necessary gravitational conversion factor, and
effective particle diameter, respectively. The void fraction is the ratio of the
interstitial volume within the packing material to the full volume of the bed.
It can be expressed by the following equation:
Ergun realized that at turbulent flow rates, viscous energy is negligible
compared to kinetic energy. Ergun postulated that there exist a smooth
transition between the domination of viscous energy losses and kinetic
energy losses. Hence, a single equation could describe an entire range of
flow rates. Ergun and Orning considered the possibility that kinetic and
viscous energy losses were additive. For non-spherical Particles, the
pressure drop can be written as follows:
Substances and equipment's used:
1-Two section column packed with identical Rashing rings of 440 m2/m3
specific surface area .
2-Pump for circulation of water .
3-Water reservoir tank .
4- Monometer .
Procedure:
1. Fill the reservoir tank with water .
2. Start the compressor and let the air pass through the column for a
period sufficient to remove all the moisture existing there in .
3. Change the air flow rate by adjusting the relevant valve and record the
pressure drop each time .
4. Stop the air flow and start the water pump for one minute or more
then stop the flow of water .
5. After a period of 2 minute, Which is sufficient for draining the water
from the column, repeat step3.
6. Repeat step 3 using different water and air flow rates . Record each
time the air flow rate at which flooding occurs.
7. Record the measurements in the table attached herewith.
Calculations :
1-Dry packed column:-
The experimental results:
∆P(N/m2)= ∆P(mm) H2O * 9.81
1- ∆P=28 (mm) H2O
∆P=28 (mm) H2O * 9.81= 274.68 N/m2
Q =180(L/min) * 1.67*10-5
Q =0.003 (m3/sec)
A (area of column) = d2 = * (0.075)2
A= 4.415*10-3 m2
Ug = = = 0.68 m/s
∆P(mm) H2O ∆P(N/m2) Q = (L/min) Q = (m3/sec) Ug = (m/s)
= ∆P(mm) H2O * 9.81
28 274.68 180 0.003006 0.680860702
28 274.68 160 0.002672 0.605209513
20 196.2 130 0.002171 0.491732729
15 147.15 80 0.001336 0.302604757
10 98.1 60 0.001002 0.226953567
6 58.86 50 0.000835 0.189127973
4 39.24 40 0.000668 0.151302378
3 29.43 30 0.000501 0.113476784
3 29.43 20 0.000334 0.075651189
Theoretical results:
∆P(dry)(N/m2)=L [150 *μ *ug +1.75 ]
Where: L=1.4m , μ =1.8* kg/m.s
e = 0.6 , ρ =1.2 kg/m3
1- Ug = 0.68 m/s at Q=180 L/min
∆P(dry)(N/m2)=1.4 [150 *1.84* * 0.68 +1.75 ]
∆P(dry)= 33.998 (N/m2)
Q Ug
(L/min) (m/s) ∆P(dry)(N/m2) = L [150 *μ *ug +1.75 ]
180 0.680860702 33.99829102
160 0.605209513 26.8970644
130 0.491732729 17.80318009
80 0.302604757 6.80125475
60 0.226953567 3.854576507
50 0.189127973 2.692828552
40 0.151302378 1.73880799
30 0.113476784 0.99251487
20 0.075651189 0.453949154
300
250
200
(P(exp∆
150
P(exp)
100
50
0
0 0.2 0.4 0.6 0.8
Ug
40
35
30
(P(theoratecal∆
25
20
15 Ptheor
10
0
0 0.2 0.4 0.6 0.8
Ug
2-Wet packed column:-
The experimental results:
1- At Q(air flow rate)= 30 L/min
∆P=15 (mm) H2O = 147.15 (N/m2)
Ug= 0.1134 (m/s)
Q(water flow rate) = 0.02 (L/s)
Q(air flow rate) ∆P Ug
(L/min) (N/m2) ( m/s)
30 147.15 0.113476784
40 147.15 0.151302378
50 186.39 0.189127973
60 235.44 0.226953567
70 255.06 0.264779162
80 284.49 0.302604757
90 333.54 0.340430351
100 372.78 0.378255946
110 421.83 0.41608154
120 539.55 0.453907135
130 588.6 0.491732729
140 637.65 0.529558324
150 667.08 0.567383918
160 686.7 0.605209513
170 725.94 0.643035108
Q(water flow rate) = 0.03 (L/s)
Q(air flow rate) ∆P Ug
(L/min) (N/m2) ( m/s)
30 117.72 0.113476784
40 127.53 0.151302378
50 147.15 0.189127973
60 166.77 0.226953567
70 196.2 0.264779162
80 264.87 0.302604757
90 392.4 0.340430351
100 519.93 0.378255946
110 833.85 0.41608154
120 1177.2 0.453907135
130(flooding) 2550.6 0.491732729
Q(water flow rate) = 0.044 (L/s)
Q(air flow rate) ∆P Ug
(L/min) (N/m3) ( m/s)
30 392.4 0.113476784
40 441.45 0.151302378
50 470.88 0.189127973
60 549.36 0.226953567
70 598.41 0.264779162
80 627.84 0.302604757
90 667.08 0.340430351
100 833.85 0.378255946
110 1373.4 0.41608154
120(flooding) 1814.85 0.453907135
Q(water flow rate) = 0.07(L/s)
Q(air flow rate) ∆P Ug
(L/min) (N/m2) ( m/s)
30 441.45 0.113476784
40 480.69 0.151302378
50 588.6 0.189127973
60(flooding) 627.84 0.226953567
Theoretical results:
1-∆P(wet)= ∆P(dry) * (1+ )
∆P(dry)= 33.99829102 ( N/m2)
∆P(wet)= 33.99829102 * (1+ )
∆P(wet)= 12500.03833 (N/m2)
∆P(dry) ∆P(wet) N/m
2
= ∆P(dry) * (1+ )
2
N/m
33.99829102 12500.03833
26.8970644 9889.154011
17.80318009 6545.63588
6.80125475 2500.594663
3.854576507 1417.199296
2.692828552 990.0632976
1.73880799 639.3017377
0.99251487 364.9146339
0.453949154 166.9019723
800
700
600
500
(P(exp∆
400
300 Pexp
200
100
0
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7
Ug
Water flow rate =0.02
3000
2500
2000
(P(exp∆
1500
Pexp
1000
500
0
0 0.1 0.2 0.3 0.4 0.5 0.6
Ug
Water flow rate = 0.03
2000
1800
1600
1400
1200
∆P(exp)
1000
800 Pexp
600
400
200
0
0 0.1 0.2 0.3 0.4 0.5
Ug
Water flow rate = 0.044
700
600
500
400
∆P(exp)
300
Pexp
200
100
0
0 0.05 0.1 0.15 0.2 0.25
Ug
Water flow rate = 0.07
14000
12000
10000
8000
∆P(wet)
6000 Pwet
4000
2000
0
0 0.2 0.4 0.6 0.8
Ug
Discussion:
When comparing Ergun’s equation to the experimental data, it is seen
that the Ergun equation deviates from the experimental data. The
factors that can be attributed to this deviation could be due to the
assumptions made about the apparatus. When deriving this equation,
Ergun used packing material with a rough surface whereas in this
experiment the packing materials were rather smooth .
As can be seen when increase the flow of air that causes pay the water
towards the top where at a certain value the flooding phenomenon is
occurring .
