Waste water treatment: Settling

Settling of Particles Settling Types Type I: Horizontal, discrete, or free settling Examples are plain settling in water treatment plants and grit settling in wastewater treatment plants Type II: Settling of flocculent particles in a dilute suspension: Examples are settling of flocs after coagulation and flocculation in water and wastewater treatment plants and grit primary settling in wastewater treatment plants Type III: Zone settling or hindered settling Example is the settling of intermediate concentration of particles where hindernece in the settling occurs because the particles are close from each other This occurs in the secondary clarifiers at wastewater treatment plants Type IV: Compression settling where the particles are too close that settling can occur by compaction This occurs in secondary clarifiers in wastewater treatment plants when the solids at the bottom are thickened From MIT OCW Type II Type I Influent GC Type III & IV PST AT SST TF BS ST AD Effluent CT SF BS= Bar screens, GC= Grit chamber, PST= Primary settling tank, AT= Aeration tank, SST= Secondary settling tank, TF= Trickling filter, SS= settling tank, SF= Sand filter, CT= chlorination tank, AD= Anaerobic digester Type I: Horizontal, discrete, or free settling The settling velocity (vs) of discrete particles is given by Stokes Law: g (ρ s − ρ)d vs = 18µ Where, ρs and ρw are the particle and water density, respectively, d is the particle diameter, and µ is water absolute viscosity= 1.01 g/(m.sec) Once vs is determined, the over-flow rate vo is set to be equal to vs in an ideal reactor However, since reactors deviate from ideality, v o is set at 0.33-0.7 times v Example m3/s 0.15 Will grit particles of 0.2 mm or larger be collected in a horizontal grit chamber that is 13.5 m in length? The design flow through the grit chamber is 0.15 m3/s, the width of the chamber is 0.56 m, and the horizontal velocity is 0.25 m/s? h vx= 0.25 m/s vs 13.5m Solution g ( ρ s − ρ )d 9.81(2650 − 1000)(0.2 ×10 −3 ) vs = = = 0.00356 m / s −3 18µ 18 × (1.01×10 ) Compare the time it takes the particle to reach the bottom (t s) with the time the water carrying the particle remains in the reactor (t h) Ax = Q/vx = 0.15/0.25 = 0.6 m2 0.6 = h*0.56 h= 1.07 m ts= h/vs = 1.07/0.0356 = 30.1 sec th = V/Q = 0.6*13.5/0.15 = 54 sec Since th > ts, then the particle will be collected Type II: Settling of flocculent particles in a dilute suspension cm 13-20.5 The design of these settling tanks usually requires a batch laboratory test using a column with a depth similar to that in the treatment plant (> 3m) and a diameter of 130-205 mm After pouring the waste in the column, samples are taken at different depths for different times The samples are tested for 3m> suspended solids and the results are reported as percentage of the initial suspended solids Lines of equal percentage of suspended solids are then drawn as shown in the figure below Sampling ports Example The settlement of flocculent particles after coagulation of a surface water with suspended solids concentration of 200 mg/l was determined by a settling column test The concentration of suspended solids at different depths and times are listed below Sketch lines representing 50%, 60%, and 70% removals For a depth of 4.0 m, estimate the overall removal for a settling time of 120 (t (min m 1.0 m 2.0 m 3.0 m 4.0 20 104 120 134 138 40 78 102 114 122 60 70 90 94 96 80 64 82 86 94 100 60 70 76 84 120 52 64 70 78 150 50 58 66 74 1m 2m 3m 4m ( Suspended solids concentration (C Solution : Find the percentage removal at each depth and time using Percentage removal= (Co-C/Co)*100 (t (min Percentage removal m 1.0 m 2.0 m 3.0 m 4.0 20 48 40 33 31 40 61 49 43 39 60 65 55 53 52 80 68 59 57 53 100 70 65 62 58 120 74 68 65 61 150 75 71 67 63 Insert percent removal on a time-depth chart Depth, m 48 61 65 68 70 74 75 40 49 55 59 65 68 71 33 43 53 57 62 65 67 31 20 39 40 52 60 53 80 58 100 61 120 Time, 63 140 150 Draw iso-concentration lines 100% 48 61 65 68 70 74 70% 60% Depth 40 49 75 55 59 65 68 71 67 50% 33 43 53 57 62 65 31 20 39 40 52 60 53 80 58 100 61 120 Time 63 140 150 Determine the percent removal (R) at the time needed 100% Z2=0.8 70% 60% Depth 50% Z1=2.7 20 40 60 80 Time R 120 = 61 + (70 − 61)( 58 100 61 120 63 140 150 This line corresponds to 61% 2.7 0.8 ) + (100 − 70)( ) = 73% 4 Design detention time By finding Rt at different times, one can then plot Rt versus t From this graph one can select the detention time needed for a certain percent removal This value is multiplied by 1.75 for scale-up purposes Rt Desired removal Needed detention time t Type III: Zone settling or hindered settling Type IV: Compression settling For design: Batch settling tests are usually done in a 1-liter graduate cylinder A settling curve (clarified depth versus time) is generated as shown below 0.4 40cm cylinder t=0 t1 0.2 , m interface height 0.3 0.1 t2 0 10 20 30 40 time, 50 60 70 80 0.4 , m 0.2 0.1 Determine the area required for clarification: Ac= Q/ vo 10 20 30 40 50 60 70 80 50 60 70 80 time, 0.4 0.3 0.2 , m interface height Extend the tangents from the hindered settling region and the compression region and bisect the angle formed to locate point Draw a tangent to the curve at point Know the initial concentration Co and the initial height Ho, select a design underflow concentration Cu and determine the interface height Hu using CoHo= CuHu Draw a horizontal line from Hu to intersect the tangent line and determine the time tu This is the time required to reach Cu Determine the area required for thickening: At= Q×tu/Ho vo=slope 0.3 interface height The area of the final clarifier is then obtained from the curve as follows: Determine the slope of the hindered settling region, vo This is the settling velocity required for clarification 0.1 Hu tu 0 10 20 30 40 time, Example A final clarifier is to be designed for an activated sludge plant treating industrial wastewater having a design flow of 5000 m 3/d Batch settling studies have been conducted in the laboratory using acclimated culture of activated sludge (Co=2,500 mg/L) and a graduated cylinder (0.4m height) The settling curve is as shown below If the design underflow concentration (Cu) is 12,000 mg/L What is the diameter of the settling tank? 0.4 0.2 , m interface height 0.3 0.1 0 10 20 30 40 time, 50 60 70 80 Solution Determine the slope of the hindered settling zone From the graph the slope is 0.015 This is the overflow rate vo (5000)( ) Q 60 × 24 = 231.5m = Area for clarification (Ac)= vo 0.015 Determine Hu from Co H o = Cu H u ⇒ H u = 2500 × 0.4 = 0.083m 12000 Determine the time (tu) needed to thicken the sludge to a depth of Hu Following the procedure described (see next figure) we have tu=35 Thus, the area needed for thickening (At) = Qtu = Ho (5000)( )(35) 24 × 60 = 304m 0.4 The design area is 304m2 Thus, the diameter of the tank is 19.7 m 0.4 vo=slope=0.015 m/min 0.2 , m interface height 0.3 0.1 Hu tu=35 0 10 20 30 40 time, 50 60 70 80 [...]... percent removal This value is multiplied by 1.75 for scale-up purposes Rt Desired removal Needed detention time t Type III: Zone settling or hindered settling Type IV: Compression settling For design: Batch settling tests are usually done in a 1-liter graduate cylinder A settling curve (clarified depth versus time) is generated as shown below 0.4 40cm cylinder t=0 t1 0.2 , m interface height 0.3 0.1... obtained from the curve as follows: 1 Determine the slope of the hindered settling region, vo This is the settling velocity required for clarification 0.1 Hu tu 0 0 10 20 30 40 time, min Example A final clarifier is to be designed for an activated sludge plant treating industrial wastewater having a design flow of 5000 m 3/d Batch settling studies have been conducted in the laboratory using acclimated... activated sludge (Co=2,500 mg/L) and a graduated cylinder (0.4m height) The settling curve is as shown below If the design underflow concentration (Cu) is 12,000 mg/L What is the diameter of the settling tank? 0.4 0.2 , m interface height 0.3 0.1 0 0 10 20 30 40 time, min 50 60 70 80 Solution Determine the slope of the hindered settling zone From the graph the slope is 0.015 This is the overflow rate... 0.4 , m 0.2 0.1 0 2 Determine the area required for clarification: Ac= Q/ vo 0 10 20 30 40 50 60 70 80 50 60 70 80 time, min 0.4 0.3 0.2 1 , m interface height 3 Extend the tangents from the hindered settling region and the compression region and bisect the angle formed to locate point 1 4 Draw a tangent to the curve at point 1 5 Know the initial concentration Co and the initial height Ho, select a