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 (ππ/π3 ) 0.016706 0.016706 0.016706 0.016706 0.016706 Slurry Concentration is time Volume Total Time Increment Increment Volume βπ (bar) (s) (s) (L) (L) βπ‘/βπ 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 π€ is calculated as follows: π€= πππ π πΆππΆπ3 1 ππ × = 0.016706 3 ππππ’ππ ππ π€ππ‘ππ 1000 π The pressure taken is 1 bar, where the time and volume are measured and represented in the table. The determination of βπ‘/βπ allows the creation of a graph that helps in the analysis. The graph represented below shows the relationship between βπ‘/βπ and π. The equation of the graph is taken from the excel sheet, that is used to continue the calculation. 60 βπ‘/βπ (S/L) 50 y = 6,9444x - 11,389 40 30 20 10 0 0 2 4 6 8 10 V (L) Figure 2 βt/βV 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 βt/βV against total V at constant βP 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 π¦ = 0: 0 = 6.9444π₯ − 11.389 π₯ = 11.389/6.9444 = 1.64 Thus, the value of the imaginary volume, V’, is equal to 1.64 πΏ which is equal to 0.00164 π3. Estimation of πΌ, the Specific cake resistance: Depending on the equation of the flow rate and its derivation, as shown in the equation below: ππ π π 3 × βπ = ππ‘ = π£ = π΄ π΄ 5 × μ × π 2 × (1 − π)2 × πΏ This equation is Carman-Kozeny equation, and substituting the values of L and πΌ in this equation gives the derivative equation as follows: βπ‘ (π + π ′ )(πΌ × μ × w) = βπ π΄2 × βπ The Cross-sectional area of filter cake is determined by: π΄ = πΏ × π = 0.18 × 0.17 = 0.0306 π2 π΄2 = 0.00093636 π4 Rearranging the equation of βπ‘ βπ gives the following form: βπ‘ × π΄2 × βπ βπ (πΌ × μ × w) = (π + π ′ ) A single point from the lab outcomes is taken and the table below shows the values: Table 1 Lab point βπ‘ π ( ) βπ π 51.66 βπ‘ π ( ) βπ π3 51666.66 π(πΏ) π(π3 ) π ′ (πΏ) π ′ (π3 ) 9 0.009 1.64 0.00164 Substituting these values in the above equation gives the value of (πΌ × μ × w). βπ‘ × π΄2 × βπ 51666.66 × 0.00093636 × 100000 βπ (πΌ × μ × w) = = = 454684958 (π + π ′ ) 0.009 + 0.00164 The viscosity of the solvent, water, is equal to 10−3 ππ. π πΌ= 454684958 454684958 π = −3 = 2.72 × 1013 μ×w 10 × 0.016706 ππ After determining the Specific cake resistance, the Specific Surface of particles is determined using the equation of πΌ as follows: 5 × π 2 (1 − π) πΌ= ππ × π 3 The density of the solid, CaCO3 is given 2800 ππ/π3 . The Cake voidage, π, is determined using the corresponding equation: π= ππ − ππ ππ 932.75 =1− =1− = 0.66 ππ ππ 2800 Bulk density of bed, ππ , is calculated as well depending on the corresponding equation: ππ = πππ π ππ π‘βπ ππππ 0.2095 ππ = = 932.75 3 π£πππ’ππ ππ π‘βπ ππππ 0.00022 π π£πππ’ππ ππ π‘βπ ππππ = πΏ × π × π‘βππππππ π = 0.18 × 0.17 × 0.00734 = 0.00022 π3 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 πΌ, gives the following shape: π2 = πΌ × ππ × π 3 2.72 × 1013 × 2800 × 0.663 = = 2.266 × 1011 5 × (1 − π) 5 × (1 − 0.66) π = √2.266 × 1011 = 4.76 × 105 π−1 The Apparent particle diameter is then determined by dividing 6 over S. π₯= 6 6 = = 1.26 × 10−5 π π 4.76 × 105 The Gradient of graph, φ, is determined then accordingly: φ= (πΌ × μ × w) 454684958 π = = 4855877.632 6 2 π΄ × βπ 0.00093636 × 100000 π After determining the Gradient of graph, the constants of the integrated equation can be determined. π‘ × βπ = ππ + ππ 2 a represents the filter cloth and b the filter cake resistance. π = φ × V ′ × βπ = 4855877.632 × 0.00164 × 100000 = 796376797.9 π =φ× ππ π4 . π βπ ππ = 2.42 × 1011 7 2 π .π The flow rate is estimated according to the following equation, after determining the depth of the bed. πΏ= (π + π ′ )π€ = 6.22 × 10−6 π (1 − π) × ππ × π΄ π 3 × βπ π3 π= × π΄ = 1.158 5 × μ × S 2 × (1 − π)2 × πΏ π Test 2 Data collected: βπ‘/βπ s/l time Slurry Concentration Total Time Increment (ππ/π3 ) βπ (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 βπ‘/βπ 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 βt/βV 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 (π3 ) Pb (ππ/π3 ) Ps (ππ/π3 ) Epsilon V’ (π3 ) Area (π2 ) π΄πππ2 (π4 ) W (ππ/π3 ) L (m) Q (π3 /π ) µ (pa.s) π α (ππ) 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 (π6 ) π 5300024.086 S (π−1 ) X (m) ππ a (π4 .π ) 3.99 × 105 1.5 × 10−5 2923520872 ππ 5.3 × 1011 b (π7 .π ) Comparison of the filtration rate and the specific cake resistance depending on pressure variation. βπ‘/βπ 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 (π3 /π ) Bed Depth (m) π Cake Resistance (α) (ππ) 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 π /π compared to Test 2, which was conducted at 2 bar with a flow rate of 1.36 π3 /π . 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 π/ππ. 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.