ANSWERS AREW7/T1 (a) 2.58 for suspension A and 2.49 for suspension B SHOW WORKING ON HOW TO REACH THESE PLEASE DONT COPY AND PASTE FROM OTHER ANSWERED QUESTIONS AS THESE ARE WRONG TASK 1:A suspension is divided in two different aliquots containing 24 mL of solvent phase. Eachaliquot is filtered under constant pressure conditions using different filtration conditions: slurryA is filtered using a pressure drop of 200 mbar, while slurry B is filtered using a pressure dropof 700 mbar. The diameter of the Nutsche filter is 27 mm.The slurry comprises ethanol (solvent density 0.789 g/cm³ and dynamic viscosity 1.095 CP)and spherical particles. Volume of cake formed by the passage of unit volume of filtrate isequal to 3. Filtration flow rate and time/volume of filtrate versus volume of filtrate removedduring filtration for the two slurries are reported in the tables below.Slurry AFiltrate Volume A (mL)0152224Slurry BFiltrate Volume B (mL)0122024Time/Filtrate volume (s/m³)2.5E+062.0E+061.5E+061.0E+065.0E+05From the filtration of slurries A and B, the graph of time/filtration volume is shown below. Foreach filtration a linear fit is shown.0.0E+00Filtrate Volume A (m³) Time A (s)001.5 x 10102.2 x 10-5152.4 x 10-518Filtrate Volume B (m³)01.2 x 10-50.E+002 x 10-62.4 x 10-5-Filtration A-Filtration BTime B (s)0153447Time/Volume A (s/m³)N/A6.67 x 1056.82 x 1057.5 x 105ON TENDENTime/Volume B (s/m³)N/A1.25 x 1081.7 x 1061.96 x 108CP414 Particle Technology Workshop 7y = 8.12E+10x + 9.03E+041.E-052.E-05Filtrate volume (m³)y = 3.10E+10x + 5.18E+043.E-05(a) Calculate the compressibility index, n (-).(b) Identify if the cake is incompressible, compressible or highly compressible, to understandif Darcy's law is valid for this suspension. Example: leaf filtrationLeaf area=0.5m²Constant pressure drop-500kPaIncompressible cake producedVolume of filtrate 0.1collected (m³)Time [s]Solution point 1140Constant pressure filtration:ELEesetELCalculate:1. Time required to collect 0.8m² of filtrate at constant pressure drop of 700kPa2. Time required to wash the resulting cake with 0.3m³ of water at pressure drop of 400kPa.0This worked example is reported in the suggested book "Introduction to particle technology" by Martin Rhodes.This book is available through myPlace through the library (digital version available).200Solution point 20.2360NObationicuppe24²(-6P) = 4000Ergun equation:-AP0.310001dVAdt660tcropμR40μV 24²(-AP) A²-AP). In this case you don't need the value of RmGRADIENTbecause you are just considering thegradient and the interceptA=0.5m² and (-AP) = 500 10³ Pa, app = 1 × 10° Pas/m² and V-0.125 m³So substituting values in510(4V+1) gives t-2400s (40mins)(-AP)INTERCEPTFiltration cake resistance: a =wwwZADO= 150 HU(1-8)²x²3ΔΡEuler number: Eu =Pp0²/2M500Ru4²(-AP) V-1000During filtration:• cake thickness continuously increases• The volume of flow rate of filtrate continuously decrease (constant pressure case)0(-AP)Aau(V+Veg)tConstant pressure drop filtration: ==0.41040150(1-ɛ)²£3flow rate of wash (at 400kPa) = 1.89 +10 200-101.0810 m³/sHence the time needed to pass 0.3m³ of wash solvent at this flow rate is 2778s (46.3min).Cake compressibility index: In a = n In APSubstituting the filtrate passed at the end of filtration (0.8m3), using apu = 1+ 10⁹ Pas/m², V-0.125 m³ and (-AP) = 700.10³ Pal• Gradient part of the question is related tocake resistance1.89 104 m³/sAssuming the wash solvent (water) has the same properties as the filtrate (density and viscosity), wash rate is equal to 1.8910 m³/s at (-AP) = 700 + 10³ Pa.However, during the washing the applied pressure difference is 400 10³ Pa:αφμV 24²(-AP)• Intercept part of the equation is related tomedia resistance-4000-1000Gas cyclone characteristic gas velocity: v =Va of+1.75 PfU²(1-8)XE³0.51500Correct solution:три• Divide axu by 2²(4)• Divide (4V + 1) by 2-(2V+0.5)ES-V+2APEu pfRmDuA²(-AP)DEVea

Question

ANSWERS AREW7/T1 (a) 2.58 for suspension A and 2.49 for suspension B SHOW WORKING ON HOW TO REACH THESE PLEASE DONT COPY AND PASTE FROM OTHER ANSWERED QUESTIONS AS THESE ARE WRONG TASK 1:A suspension is divided in two different aliquots containing 24 mL of solvent phase. Eachaliquot is filtered under constant pressure conditions using different filtration conditions: slurryA is filtered using a pressure drop of 200 mbar, while slurry B is filtered using a pressure dropof 700 mbar. The diameter of the Nutsche filter is 27 mm.The slurry comprises ethanol (solvent density 0.789 g/cm³ and dynamic viscosity 1.095 CP)and spherical particles. Volume of cake formed by the passage of unit volume of filtrate isequal to 3. Filtration flow rate and time/volume of filtrate versus volume of filtrate removedduring filtration for the two slurries are reported in the tables below.Slurry AFiltrate Volume A (mL)0152224Slurry BFiltrate Volume B (mL)0122024Time/Filtrate volume (s/m³)2.5E+062.0E+061.5E+061.0E+065.0E+05From the filtration of slurries A and B, the graph of time/filtration volume is shown below. Foreach filtration a linear fit is shown.0.0E+00Filtrate Volume A (m³) Time A (s)001.5 x 10102.2 x 10-5152.4 x 10-518Filtrate Volume B (m³)01.2 x 10-50.E+002 x 10-62.4 x 10-5-Filtration A-Filtration BTime B (s)0153447Time/Volume A (s/m³)N/A6.67 x 1056.82 x 1057.5 x 105ON TENDENTime/Volume B (s/m³)N/A1.25 x 1081.7 x 1061.96 x 108CP414 Particle Technology Workshop 7y = 8.12E+10x + 9.03E+041.E-052.E-05Filtrate volume (m³)y = 3.10E+10x + 5.18E+043.E-05(a) Calculate the compressibility index, n (-).(b) Identify if the cake is incompressible, compressible or highly compressible, to understandif Darcy's law is valid for this suspension. Example: leaf filtrationLeaf area=0.5m²Constant pressure drop-500kPaIncompressible cake producedVolume of filtrate 0.1collected (m³)Time [s]Solution point 1140Constant pressure filtration:ELEesetELCalculate:1. Time required to collect 0.8m² of filtrate at constant pressure drop of 700kPa2. Time required to wash the resulting cake with 0.3m³ of water at pressure drop of 400kPa.0This worked example is reported in the suggested book &#34;Introduction to particle technology&#34; by Martin Rhodes.This book is available through myPlace through the library (digital version available).200Solution point 20.2360NObationicuppe24²(-6P) = 4000Ergun equation:-AP0.310001dVAdt660tcropμR40μV 24²(-AP) A²-AP). In this case you don't need the value of RmGRADIENTbecause you are just considering thegradient and the interceptA=0.5m² and (-AP) = 500 10³ Pa, app = 1 × 10° Pas/m² and V-0.125 m³So substituting values in510(4V+1) gives t-2400s (40mins)(-AP)INTERCEPTFiltration cake resistance: a =wwwZADO= 150 HU(1-8)²x²3ΔΡEuler number: Eu =Pp0²/2M500Ru4²(-AP) V-1000During filtration:• cake thickness continuously increases• The volume of flow rate of filtrate continuously decrease (constant pressure case)0(-AP)Aau(V+Veg)tConstant pressure drop filtration: ==0.41040150(1-ɛ)²£3flow rate of wash (at 400kPa) = 1.89 +10 200-101.0810 m³/sHence the time needed to pass 0.3m³ of wash solvent at this flow rate is 2778s (46.3min).Cake compressibility index: In a = n In APSubstituting the filtrate passed at the end of filtration (0.8m3), using apu = 1+ 10⁹ Pas/m², V-0.125 m³ and (-AP) = 700.10³ Pal• Gradient part of the question is related tocake resistance1.89 104 m³/sAssuming the wash solvent (water) has the same properties as the filtrate (density and viscosity), wash rate is equal to 1.8910 m³/s at (-AP) = 700 + 10³ Pa.However, during the washing the applied pressure difference is 400 10³ Pa:αφμV 24²(-AP)• Intercept part of the equation is related tomedia resistance-4000-1000Gas cyclone characteristic gas velocity: v =Va of+1.75 PfU²(1-8)XE³0.51500Correct solution:три• Divide axu by 2²(4)• Divide (4V + 1) by 2-(2V+0.5)ES-V+2APEu pfRmDuA²(-AP)DEVea

Accepted Answer

A suspension with two different aliquotas is given Amount of solventConstant pressure filtration…