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Example: leaf filtration Leaf area=0.5m² Constant pressure drop-500kPa Incompressible cake produced Volume of filtrate 0.1 collected (m³) Time (s) 140 0.2 360 0.3 660 Pressure leaf filters http://slideplayer.com/da/4134960 0.4 1040 Calculate: 1. Time required to collect 0.8m³ of filtrate at constant pressure drop of 700kPa 2. Time required to wash the resulting cake with 0.3m³ of water at pressure drop of 400kPa. 0.5 1500

Example: leaf filtration Leaf area=0.5m² Constant pressure drop-500kPa Incompressible cake produced Volume of filtrate 0.1 collected (m³) Time (s) 140 0.2 360 0.3 660 Pressure leaf filters http://slideplayer.com/da/4134960 0.4 1040 Calculate: 1. Time required to collect 0.8m³ of filtrate at constant pressure drop of 700kPa 2. Time required to wash the resulting cake with 0.3m³ of water at pressure drop of 400kPa. 0.5 1500

Introduction to Chemical Engineering Thermodynamics
Introduction to Chemical Engineering Thermodynamics
8th Edition
ISBN:9781259696527
Author:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Publisher:J.M. Smith Termodinamica en ingenieria quimica, Hendrick C Van Ness, Michael Abbott, Mark Swihart
Chapter1: Introduction
Section: Chapter Questions
Problem 1.1P
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could someone show clearer working for 1 and 2 

 

i am not sure where the (4v+1) comes from

Example: leaf filtration Leaf area=0.5m² Constant pressure drop=500kPa Incompressible cake produced Volume of filtrate 0.1 collected (m³) Time (s) Solution point 1 Constant pressure filtration: 0.6 0.5 04 0.3 0.2 140 0.1 Calculate: 1. Time required to collect 0.8m³ of filtrate at constant pressure drop of 700kPa 2. Time required to wash the resulting cake with 0.3m³ of water at pressure drop of 400kPa. 0 0 This 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). 200 0.2 400 Solution point 2 360 600 1800 1000 1200 Filtration time (s) t αφμ V2A²(-AP) GRADIENT αφμ 2A²(-AP) 0.3 = 4000 660 1400 1600 Pressure leaf filters V + R„φμ A² (-AP) INTERCEPT 3500 3000 2500 2000 1500 1000 500 0 http://slideplayer.com/da/4134960 Vea 0.1 R„φμ A²(-AP) Veq = 1000 A=0.5m² and (-AP) = 500 10³ Pa, au = 1 10⁹ Pas/m² and Veq=0.125 m³ So substituting values in (4V + 1) gives t=2400s (40mins) 0.5*10⁰ t V (-AP) 1 dv (-AP)A A dt au(v + Veq) = 0.4 1040 During filtration: cake thickness continuously increases The volume of flow rate of filtrate continuously decrease (constant pressure case) • Gradient part of the question is related to cake resistance • Intercept part of the equation is related to media resistance y=4000x+1000 • In this case you don't need the value of Rr because you are just considering the gradient and the intercept 0.5 0.3 0.4 Volume of filtrate (m²) 1500 0.5 flow rate of wash (at 400kPa) = 1.89 10-4 (400+10³) 01.08 10-4 m³/s (700+10¹) Correct solution: • Divide au by 22 (-AP)) αφμ • Divide (4V + 1) by 2 = (2V+0.5) Hence the time needed to pass 0.3m³ of wash solvent at this flow rate is 2778s (46.3min). 0.6 Substituting the filtrate passed at the end of filtration (0.8m3), using apμ = 1 10⁹ Pas/m², Veq=0.125 m³ and (-AP) = 700 * 10³ Pa dv 1.89 10-4 m³/s * dt Assuming the wash solvent (water) has the same properties as the filtrate (density and viscosity), wash rate is equal to 1.89 * 10-4 m³/s at (-AP) = 700 10³ Pa. However, during the washing the applied pressure difference is 400 10³ pa:
Transcribed Image Text:Example: leaf filtration Leaf area=0.5m² Constant pressure drop=500kPa Incompressible cake produced Volume of filtrate 0.1 collected (m³) Time (s) Solution point 1 Constant pressure filtration: 0.6 0.5 04 0.3 0.2 140 0.1 Calculate: 1. Time required to collect 0.8m³ of filtrate at constant pressure drop of 700kPa 2. Time required to wash the resulting cake with 0.3m³ of water at pressure drop of 400kPa. 0 0 This 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). 200 0.2 400 Solution point 2 360 600 1800 1000 1200 Filtration time (s) t αφμ V2A²(-AP) GRADIENT αφμ 2A²(-AP) 0.3 = 4000 660 1400 1600 Pressure leaf filters V + R„φμ A² (-AP) INTERCEPT 3500 3000 2500 2000 1500 1000 500 0 http://slideplayer.com/da/4134960 Vea 0.1 R„φμ A²(-AP) Veq = 1000 A=0.5m² and (-AP) = 500 10³ Pa, au = 1 10⁹ Pas/m² and Veq=0.125 m³ So substituting values in (4V + 1) gives t=2400s (40mins) 0.5*10⁰ t V (-AP) 1 dv (-AP)A A dt au(v + Veq) = 0.4 1040 During filtration: cake thickness continuously increases The volume of flow rate of filtrate continuously decrease (constant pressure case) • Gradient part of the question is related to cake resistance • Intercept part of the equation is related to media resistance y=4000x+1000 • In this case you don't need the value of Rr because you are just considering the gradient and the intercept 0.5 0.3 0.4 Volume of filtrate (m²) 1500 0.5 flow rate of wash (at 400kPa) = 1.89 10-4 (400+10³) 01.08 10-4 m³/s (700+10¹) Correct solution: • Divide au by 22 (-AP)) αφμ • Divide (4V + 1) by 2 = (2V+0.5) Hence the time needed to pass 0.3m³ of wash solvent at this flow rate is 2778s (46.3min). 0.6 Substituting the filtrate passed at the end of filtration (0.8m3), using apμ = 1 10⁹ Pas/m², Veq=0.125 m³ and (-AP) = 700 * 10³ Pa dv 1.89 10-4 m³/s * dt Assuming the wash solvent (water) has the same properties as the filtrate (density and viscosity), wash rate is equal to 1.89 * 10-4 m³/s at (-AP) = 700 10³ Pa. However, during the washing the applied pressure difference is 400 10³ pa:
−ΔΡ Ergun equation: = 150 μU(1-⁹)² x²x3 150(1-)² xv 23 Cake compressibility index: In a = n ln AP Filtration cake resistance: a = Constant pressure drop filtration: +1.75 PƒU² (1−ɛ) xɛ3 V αφμ 2A²(-AP) ·V + RmQμ A²(-AP) Vea
Transcribed Image Text:−ΔΡ Ergun equation: = 150 μU(1-⁹)² x²x3 150(1-)² xv 23 Cake compressibility index: In a = n ln AP Filtration cake resistance: a = Constant pressure drop filtration: +1.75 PƒU² (1−ɛ) xɛ3 V αφμ 2A²(-AP) ·V + RmQμ A²(-AP) Vea
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could someone show clearer working for 1 and 2 please

 

i am not sure where the (4v+1) comes from

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