Separation Processes - CEE_5_SEP
Filter Press
Effect of pressure on filtration
Prepared by: Rashid Alenezi (4106937)
Introduction
The filter press functions by applying pressure to facilitate the cake filtering process, whereby a slurry is passed
through the press containing filtration meshes. The purpose of this study was to examine the correlation between
varying concentrations and the resulting weight and volume of cake generated in a filter press. The study also
intended to assess the influence of concentration on the duration needed to get a certain amount of filtrate, as
well as to graphically depict the correlation between the volume of filtrate and the time necessary per unit
volume. Two experiments took place under varying pressure conditions, Test 1 at a pressure of 1 bar and Test 2
at a pressure of 2 bar. The obtained data exhibited discrepancies in the flow rate, bed depth, and cake resistance
observed throughout the two tests. In Test 1, there was a little increase in flow rate and a decrease in cake
resistance.
Conversely, Test 2 had a reduced bed depth but slightly greater cake resistance. The experiment further included
the determination of specific cake resistance, apparent particle diameter, and several other characteristics to
assess the filtering procedure. The study's findings revealed the inherent trade-offs among pressure, flow rate,
and cake resistance, offering significant insights into the practical implications of filter press use.
Figure 1 Process Flow Diagram of Filter Press
Test 1
Data Collected
Slurry Concentration
(𝑘𝑔/𝑚3 )
0.016706
0.016706
0.016706
0.016706
0.016706
Slurry Concentration is
time
Volume
Total Time Increment
Increment
Volume
∆𝑃 (bar)
(s)
(s)
(L)
(L)
∆𝑡/∆𝑉
1
15
15
1.8
1.8
8.333333
1
30
15
1.8
3.6
8.333333
1
58
28
1.8
5.4
15.55556
1
142
84
1.8
7.2
46.66667
1
235
93
1.8
9
51.66667
determined by dividing the mass of CaCO3 over the volume of water, or solvent. The
volume of water taken is 15 L and the mass of CaCO3 is equal to 250.59g.
Slurry concentration or 𝑤 is calculated as follows:
𝑤=
𝑚𝑎𝑠𝑠 𝐶𝑎𝐶𝑂3
1
𝑘𝑔
×
= 0.016706 3
𝑉𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 1000
𝑚
The pressure taken is 1 bar, where the time and volume are measured and represented in the table. The
determination of ∆𝑡/∆𝑉 allows the creation of a graph that helps in the analysis.
The graph represented below shows the relationship between ∆𝑡/∆𝑉 and 𝑉. The equation of the graph is taken
from the excel sheet, that is used to continue the calculation.
60
∆𝑡/∆𝑉 (S/L)
50
y = 6,9444x - 11,389
40
30
20
10
0
0
2
4
6
8
10
V (L)
Figure 2 ∆t/∆V against total V, Test 1
The initial step is the determination of the imaginary Volume, V’, that is the intersection of the graph with the Xaxis.
Figure 3 ∆t/∆V against total V at constant ∆P
The data collected is correspondingly not accurate, because the graphical representation must be a straight line as
shown in figure 2. To be able to find V’, a systematic line provided by the excel sheet is taken and its equation is
presented in the graph.
To determine V’, the prolongation of the line meets the X-axis; thus, the equation of the line is taken at y=0.
At 𝑦 = 0:
0 = 6.9444𝑥 − 11.389
𝑥 = 11.389/6.9444 = 1.64
Thus, the value of the imaginary volume, V’, is equal to 1.64 𝐿 which is equal to 0.00164 𝑚3.
Estimation of 𝛼, the Specific cake resistance:
Depending on the equation of the flow rate and its derivation, as shown in the equation below:
𝑑𝑉
𝑄
𝜀 3 × ∆𝑃
= 𝑑𝑡 = 𝑣 =
𝐴
𝐴
5 × μ × 𝑆 2 × (1 − 𝜀)2 × 𝐿
This equation is Carman-Kozeny equation, and substituting the values of L and 𝛼 in this equation gives the
derivative equation as follows:
∆𝑡 (𝑉 + 𝑉 ′ )(𝛼 × μ × w)
=
∆𝑉
𝐴2 × ∆𝑃
The Cross-sectional area of filter cake is determined by:
𝐴 = 𝐿 × 𝑊 = 0.18 × 0.17 = 0.0306 𝑚2
𝐴2 = 0.00093636 𝑚4
Rearranging the equation of
∆𝑡
∆𝑉
gives the following form:
∆𝑡
× 𝐴2 × ∆𝑃
∆𝑉
(𝛼 × μ × w) =
(𝑉 + 𝑉 ′ )
A single point from the lab outcomes is taken and the table below shows the values:
Table 1 Lab point
∆𝑡 𝑠
( )
∆𝑉 𝑙
51.66
∆𝑡 𝑠
( )
∆𝑉 𝑚3
51666.66
𝑉(𝐿)
𝑉(𝑚3 )
𝑉 ′ (𝐿)
𝑉 ′ (𝑚3 )
9
0.009
1.64
0.00164
Substituting these values in the above equation gives the value of (𝛼 × μ × w).
∆𝑡
× 𝐴2 × ∆𝑃 51666.66 × 0.00093636 × 100000
∆𝑉
(𝛼 × μ × w) =
=
= 454684958
(𝑉 + 𝑉 ′ )
0.009 + 0.00164
The viscosity of the solvent, water, is equal to 10−3 𝑝𝑎. 𝑠
𝛼=
454684958
454684958
𝑚
= −3
= 2.72 × 1013
μ×w
10 × 0.016706
𝑘𝑔
After determining the Specific cake resistance, the Specific Surface of particles is determined using the equation
of 𝛼 as follows:
5 × 𝑆 2 (1 − 𝜀)
𝛼=
𝜌𝑆 × 𝜀 3
The density of the solid, CaCO3 is given 2800 𝑘𝑔/𝑚3 .
The Cake voidage, 𝜀, is determined using the corresponding equation:
𝜀=
𝜌𝑆 − 𝜌𝑏
𝜌𝑏
932.75
=1−
=1−
= 0.66
𝜌𝑆
𝜌𝑆
2800
Bulk density of bed, 𝜌𝑏 , is calculated as well depending on the corresponding equation:
𝜌𝑏 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑘𝑒
0.2095
𝑘𝑔
=
= 932.75 3
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑘𝑒 0.00022
𝑚
𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡ℎ𝑒 𝑐𝑎𝑘𝑒 = 𝐿 × 𝑊 × 𝑡ℎ𝑖𝑐𝑘𝑛𝑒𝑠𝑠 = 0.18 × 0.17 × 0.00734 = 0.00022 𝑚3
The mass of the cake is measured to be equal to 209.5 g which is 0.2095 kg.
Substitute and re-arrange the equation of 𝛼, gives the following shape:
𝑆2 =
𝛼 × 𝜌𝑆 × 𝜀 3 2.72 × 1013 × 2800 × 0.663
=
= 2.266 × 1011
5 × (1 − 𝜀)
5 × (1 − 0.66)
𝑆 = √2.266 × 1011 = 4.76 × 105 𝑚−1
The Apparent particle diameter is then determined by dividing 6 over S.
𝑥=
6
6
=
= 1.26 × 10−5 𝑚
𝑆 4.76 × 105
The Gradient of graph, φ, is determined then accordingly:
φ=
(𝛼 × μ × w)
454684958
𝑠
=
= 4855877.632 6
2
𝐴 × ∆𝑃
0.00093636 × 100000
𝑚
After determining the Gradient of graph, the constants of the integrated equation can be determined.
𝑡 × ∆𝑃 = 𝑎𝑉 + 𝑏𝑉 2
a represents the filter cloth and b the filter cake resistance.
𝑎 = φ × V ′ × ∆𝑃 = 4855877.632 × 0.00164 × 100000 = 796376797.9
𝑏 =φ×
𝑘𝑔
𝑚4 . 𝑠
∆𝑃
𝑘𝑔
= 2.42 × 1011 7
2
𝑚 .𝑠
The flow rate is estimated according to the following equation, after determining the depth of the bed.
𝐿=
(𝑉 + 𝑉 ′ )𝑤
= 6.22 × 10−6 𝑚
(1 − 𝜀) × 𝜌𝑆 × 𝐴
𝜀 3 × ∆𝑃
𝑚3
𝑄=
× 𝐴 = 1.158
5 × μ × S 2 × (1 − 𝜀)2 × 𝐿
𝑠
Test 2
Data collected:
∆𝑡/∆𝑉 s/l
time
Slurry
Concentration
Total Time Increment
(𝑘𝑔/𝑚3 )
∆𝑃 (bar)
(s)
(s)
0.016671
2
34
34
0.016671
2
118
84
0.016671
2
205
87
0.016671
2
300
95
0.016671
2
354
59
The graphical representation of ∆𝑡/∆𝑉 vs V is shown as follows:
70
60
50
40
30
20
10
0
Volume
Increment
(L)
1.8
1.8
1.8
1.8
1.8
Volume
(L)
1.8
3.6
5.4
7.2
9
y = 5,7407x + 15,833
0
2
4
6
8
Volume (L)
Figure 4 ∆t/∆V against total V, Test 2
The same procedure of calculation is followed, and the results are tabulated as follows:
Mass of cake (kg)
Length of the cake (m)
Width of the cake (m)
Thickness of the cake (m)
Volume of the Cake (𝑚3 )
Pb (𝑘𝑔/𝑚3 )
Ps (𝑘𝑔/𝑚3 )
Epsilon
V’ (𝑚3 )
Area (𝑚2 )
𝐴𝑟𝑒𝑎2 (𝑚4 )
W (𝑘𝑔/𝑚3 )
L (m)
Q (𝑚3 /𝑠)
µ (pa.s)
𝑚
α (𝑘𝑔)
0.21439
0.17
0.175
0.00623
0.000185
1156.72
2800
0.58
2.75
0.036
0.00132
0.016671
3.937 × 10−6
1.36
10−3
8.44 × 1013
Φ (𝑚6 )
𝑠
5300024.086
S (𝑚−1 )
X (m)
𝑘𝑔
a (𝑚4 .𝑠)
3.99 × 105
1.5 × 10−5
2923520872
𝑘𝑔
5.3 × 1011
b (𝑚7 .𝑠)
Comparison of the filtration rate and the specific cake resistance depending on pressure variation.
∆𝑡/∆𝑉
8.333333
8.333333
15.55556
46.66667
51.66667
The corresponding table shows the values of the calculated cake resistance, bed depth and the flow rate at the
two different pressures.
Pressure (bar)
Flow rate (𝑚3 /𝑠)
Bed Depth (m)
𝑚
Cake Resistance (α) (𝑘𝑔)
Test 1
1
1.15
6.227 × 10−6
2.72 × 1013
−6
Test 2
2
1.36
3.93 × 10
8.446 × 1013
3
Test 1, conducted at a pressure of 1 bar, exhibits a slightly lower flow rate of 1.15 𝑚 /𝑠 compared to Test 2,
which was conducted at 2 bar with a flow rate of 1.36 𝑚3 /𝑠. Additionally, Test 1 features a greater bed depth of
6.227 × 10−6 m compared to Test 2's 3.93 × 10−6 m. However, Test 2 demonstrates a significantly higher cake
resistance (α) of 8.446 × 1013 m/kg, whereas Test 1 has a lower cake resistance of 2.72 × 1013 𝑚/𝑘𝑔.
Conclusion
In conclusion, the experiment provided valuable insights into the filtration process in a filter press and the impact
of pressure variations on key parameters. Test 1, conducted at 1 bar, demonstrated a relatively lower flow rate
and a lower cake resistance, indicating a slower filtration process than in test 2.