Report

Flocculation Monroe L. Weber-Shirk School of Civil and Environmental Engineering Overview Flocculation definition Types of flocculators Mechanical Design Fractal Flocculation Theory applied to hydraulic flocculators CFD analysis of hydraulic flocculators Hydraulic Flocculator Design Terminal velocity (mm/s) 10 1 0.1 0.01 0.001 4 110 0.001 0.01 0.1 Floc diameter (mm) 1 What is Flocculation? A process that transforms a turbid suspension of tiny particles into a turbid suspension of big particles! Requires Sticky particles (splattered with adhesive nanoglobs) Successful collisions between particles Flocs are fractals (“the same from near as from far”) The goal of flocculation is to reduce the number of colloids (that haven’t been flocculated) One goal is to understand why some colloids are always left behind (turbidity after sedimentation) Another goal is to learn how to design a flocculator Mechanical Flocculation Shear provided by turbulence created by gentle stirring Turbulence also keeps large flocs from settling so they can grow even larger! Presumed advantage is that energy dissipation rate can be varied independent of flow rate Disadvantage is the reactors have potential short circuiting (some fluid moves quickly from inlet to outlet) Recommended G and Gq values: Turbidity or Color Removal (Mechanical flocculators) Type Low turbidity, color removal High turbidity, solids removal “Velocity gradient” (G) (1/s) 20-70 70-180 Energy q* Dissipation Rate mW + (e) kg Gq (minute) 0.4 – 4.9 50,000- 11 - 210 250,000 4.9 - 32 80,000- 7 - 45 190,000 * Calculated based on G and Gq guidelines + average value assuming viscosity is 1 mm2/s Sincero and Sincero, 1996 Environmental Engineering: A Design Approach G is the wrong parameter… (Cleasby, 1984) extra Mechanical Design: mixing with paddles 3 G CD 1.9 CD AP a V 2V 3 pa extra Mechanical Flocculators? Waste (a small amount of) electricity Require unnecessary mechanical components Have a wide distribution of energy dissipation rates (highest near the paddles) that may break flocs Have a wide distribution of particle residence times (completely mixed flow reactors) Ten State Standards http://10statesstandards.com/waterrev2012.pdf The detention time for floc formation should be at least 30 minutes with consideration to using tapered (i.e., diminishing velocity gradient) flocculation. The flow-through velocity should be not less than 0.5 nor greater than 1.5 feet per minute. Agitators shall be driven by variable speed drives with the peripheral speed of paddles ranging from 0.5 to 3.0 feet per second. External, non-submerged motors are preferred. Flocculation and sedimentation basins shall be as close together as possible. The velocity of flocculated water through pipes or conduits to settling basins shall be not less than 0.5 nor greater than 1.5 feet per second. Allowances must be made to minimize turbulence at bends and changes in direction. Baffling may be used to provide for flocculation in small plants only after consultation with the reviewing authority. The design should be such that the velocities and flows noted above will be maintained. Hydraulic flocculators allowed only by special permission! Hydraulic Flocculators Horizontal baffle Vertical baffle Pipe flow Very low flows and pilot plants Gravel bed A bad idea (cleaning would require a lot of work) extra Flocculator Geometry Entrance 2 Channels Port between channels Exit Lower Baffles Upper Baffles Why aren’t hydraulic flocculators used more often? extra Simple construction means that there aren’t any items that private companies (venders) can sell as specialized components Consulting firms want to be able to pass the design responsibility off to a vender The presumed operation flexibility of mechanical flocculators (variable speed motor driving a slow mixing unit) Poor documentation of design approach for hydraulic flocculators (special permission required to use in the US!) Using electricity is cool, design innovation is suspect… Prior to AguaClara we didn’t have a design algorithm based on the fundamental physics Schulz and Okun (on Hydraulic flocculators) Surface Water Treatment for Communities in Developing Countries, (1984) by Christopher R. Schulz and Daniel A. Okun. Intermediate Technology Publications Recommend velocity between 0.1 and 0.3 m/s Distance between baffles at least 45 cm Water must be at least 1 m deep Q must be greater than 10,000 m3/day (115 L/s) These aren’t universal constants! We need to understand the real constraints so we can scale the designs correctly What length scale could make a dimensionless parameter? extra AguaClara Flocculator Design Evolution We began using conventional guidelines based on “velocity gradient” but we were aware that this system was fundamentally incorrect (it doesn’t capture the physics of turbulence) We were concerned that by using a defective model we could potentially produce defective designs If the model doesn’t capture the physics, then it won’t scale correctly (and we are designing plants for scales that hadn’t been tested) extra Edge of Knowledge Alert Why would we ever think that the baffled flocculators invented over 100 years ago were the optimal design for flocculation? We have better coagulants, shouldn’t that influence flocculator design? We are only now beginning to understand the physics of fractal flocculation We are improving the design of hydraulic flocculators based on our evolving understanding of the physics of flocculation The Challenge of Flocculation We would like to know How do particles make contact to aggregate? What determines the time required for two flocs to collide? How strong are flocs? The challenge of the large changes in scale The Al(OH)3 nanoglobs begin at a scale of 100 nanometers Flocs end at a scale of 100 micrometers Hydraulic Flocculation Theory Turbulence is caused by the expansion that results when the water changes direction as it flows around each baffle Colloids and flocs are transported to collide with each other by turbulent eddies and over short distances by viscous shear. Flocs grow in size with each successful collision Colloids have a hard time attaching to large flocs* (maybe because the surface shear is too high???) Flocculation model and collision potential for reactors with flows characterized by high Peclet numbers Monroe L. Weber-Shirk a, and Leonard W. Liona *hypothesis from 2012! http://dx.doi.org/10.1016/j.watres.2010.06.026 , How do the flocs grow? Exponential growth How many sequential collisions are required to make a 1 mm particle starting from 1 mm particles? How much larger in volume is the 1 mm diameter particle? ____________ 1,000,000,000 ! Doubling Collisions 1 collision 1+1=2 2 collisions 2+2=4 3 collisions 4+4=8 4 collisions 8+8=16 n Number of original particles in the floc = 2 n What is n to obtain 1,000,000,000 = 2 ? log(1000000000) n log(2) n=30 This assumes volume is conserved! n log(1000000000) log(2) Fractal Dimensions What happens to the density of a floc as it grows larger? Floc density approaches the density of water because the floc includes water 1 d d0i DFractal 3 V V0i DFractal Number of primary particles in the floc If volume is conserved, what is DFractal? ____ 3 Fractal Flocculation Fractal geometry explains the changes in floc density, floc volume fraction, and, ultimately, sedimentation velocity as a function of floc size The fractal dimension of flocs is approximately 2.3 (based on floc measurements) 3 V V0i DFractal Floc Volume Fraction Floc The fraction of the reactor volume that is occupied by flocs For fractal dimensions less than 3 the floc volume fraction increases as floc size increases VFloc CFloc Floc VSuspension Floc dominates Use conservation of volume to estimate initial Floc C AlOH CClay CClay C Al OH Floc Clay Al (OH ) 0 3 3 Floc 0 Clay 3 AlOH 3 Floc Volume Fraction Floc d Floc0 d0 “super fluffy” flocs 1 Floc Volume Fraction Primary particle diameter (clay + coagulant) 3 DFractal Fractal dimension = 1 Fractal dimension = 2.3 Fractal dimension = 3 0.1 0.01 0.001 4 110 5 110 0.001 0.01 0.1 Floc diamet er (mm) 1 10 Dense flocs Buoyant Density of Flocs Floc H O Floc H O 2 0 2 d0 d 3 DFractal Will these flocs settle faster than the primary particles? Floc density - water density (kg/m^3) 4 110 1000 100 10 Fractal dimension = 1 Fractal dimension = 2.3 Fractal dimension = 3 1 0.1 0.001 0.01 0.1 Floc diameter (mm) 1 10 Laminar flow Floc Terminal Velocity Upflow velocity for floc blankets Terminal velocity (mm/s) 10 1 Capture velocity for AguaClara plate settlers 0.1 0.01 0.001 4 110 gd 2 Floc H 2O Vt 18 H 2O H 2O 0.001 1 mm 2 0 gd Vt 18 H 2O 0.01 0.1 Floc diameter (mm) Floc H O d H O d0 0 2 1 Why flocculation is necessary! DFractal 1 DFractal = 2.3 and d0 = 1 mm 2 shape factor for drag on flocs The model takes into account the changing density of flocs Analytical Model of the Flocculation Process The floc porosity increases with floc diameter The velocity between flocs is a function of whether the separation distance is less or greater than the Kolmogorov scale Are the two flocs in different eddies? The time required per collision is a function of the relative velocity between flocs, the average separation distance between flocs, and the floc size In the next slides we will explore how to characterize collision time for flocs We will assume that collisions occur between similar sized flocs. That assumption will need to be evaluated, but it is probably a good assumption for the initial growth of flocs. How much water is cleared (filtered) from a floc’s perspective? Volume cleared is proportional to a 2 d Floc collision area defined by a circle with diameter = sum of the floc diameters Volume cleared is proportional to time t vr Volume cleared is proportional to the relative velocity between flocs vr t 2 VCleared d Floc vr t d 2 Floc How much volume must be cleared before a collision occurs? What is the average volume of water “occupied” by a floc? 1 Need to know floc diameter 3 LSeparation d Floc (dFloc) 6 Floc And floc volume fraction (Floc) Floc 6 3 d Floc VOccupied Floc volume Suspension volume VOccupied 6 3 d Floc Floc Use dimensional analysis to get a relative velocity given a length scale laminar Viscous range vr f e , , L LK e L is separation distance Assume linear e vr L The origin of the G notation L e 3 T L2 T Re=1 vr Lf e , 2 3 1 4 turbulent Inertial range vr f e , L vr e L 1 3 If the flocculator has laminar flow, then this side doesn’t apply and the G, Gq approach applies. Summary for Particle Collisions LK e 3 1 4 inertia viscous 2 VCleared d Floc vr tc VOccupied 6 3 d Floc vr LSeparation Floc Floc d Floc tc 6Floc vr d Floc0 d0 e tc is average time per collision vr e LSeparation 1 3 3 DFractal LSeparation Is tc a function of d? Yes! 1 1 2 d Floc 6 Floc 1 6 1 tc 2 6 e 3 Floc 3 1 1 3 Floc separation distance 1 1 6 d 3 1 tc 6 e 89 Floc 9 2 Floc Successful Collision Models Number of collisions is equal to time over collision time inertia viscous Nc t tc 1 2 1 3 Time for one tc 1 6 1 2 3 6 e Floc collision Number of successful collisions N cviscous 1 1 2 e 3 4.8Floc t N c 2 1 1 6 9 d 3 1 tc 6 e 89 Floc inertial 2 Floc 1 3 e 9 5.6Floc 2 t d Floc 8 2 3 Ncviscous 4.8Floc Gt is fractional surface coverage of colloids with adhesive nanoglobs dFloc is perhaps mean floc size of flocs that are capturing colloids For completeness we should probably include a correction for hydrodynamic effects that make it difficult for non porous particles to approach closely. This may increase the time for the first few collisions when flocs aren’t very porous Relationship between laminar and turbulent 1 3 1 3 9 4.8Floc Gq 5.6Floc 2 d Floc 8 2 2 Gq 3 4.8Floc 1 9 5.6Floc 2 d Floc 8 6 3 dClay Gq 2 9 CClay VClay Clay Gq 2 9 1 3 Set the collisions equal to get the relationship The ratio of Gq to is the number of clay particles raised to the 2/9th power! Use the target settled water turbidity as the best guess for a relationship between conventional Gq and . For a turbidity of 1 NTU the number of clay particles is 150,000,000 per liter. Schulz and Okun suggest 20,000 as a minimum value of Gq equivalent to 65 m2/3. But this isn’t calculated correctly. Should take their Gq and energy dissipation rate 2 2 9 tomake C theconversion. Gq 306 Clay 9 N Clay 2 M Clay m3 LK e 3 1 4 Minimum time to grow from colloid to large floc 400 s DFractal = 2.3 dn d0 = 1 mm n DFractal log 2 d 0 e = 6 mW/kg 120 s 40 s 20 collisions to grow from 1 mm to 0.4 mm Collision time (s) 30 20 10 0 0.001 0.01 0.1 1 10 Floc diameter (mm) 1 NT U suspension 10 NT U suspensio n 100 NT U suspension tcinertial tcviscous 1 9 8 d 1 6 31 32 9 e d Floc0 6 d0 1 3 1 2 2 d 1 6 3 Floc0 6 e d0 How much time is required to produce a 0.4 mm floc? 8 DFractal 3 9 2 DFractal 2 3 Initial Floc Growth: Phase 1 (the 50% solution) Initial growth phase of flocculation can be modeled with the equations on the previous slide (summing the collision times until the maximum floc size is reached) The end of the initial growth phase is reached when a significant fraction of the flocs reach a size that can be removed by sedimentation. Observations: Collision time model The previous analysis was tracking the time required to make the first big flocs For a highly turbid suspension it may only take a few seconds to produce visible flocs This is why successful flocculation can be observed very early in a flocculator This doesn’t mean that a flocculator with a residence time of 100 s will perform well We need to track the colloids that are left behind! residual turbidity Performance is based on _______________ Remember our Goal? Introduce pC* Sloppy parlance… log removal What is the target effluent turbidity for a water treatment plant? What is pC* for a water treatment plant treating 300 NTU water? Tracking the residual turbidity: Phase 2 of Flocculation After the initial production of flocs at their maximum size the interaction of the colloids change Most of the collisions are ineffective because collisions with flocs that are maximum size apparently are useless and most flocs are their maximum size Flocculation Model: Integrating and tracking residual turbidity dCColloid kCColloid dN c dCColloid kdN c CColloid CColloid CColloid0 dCColloid k dN c CColloid CColloid ln CColloid 0 kN c The change in colloid concentration with respect to the potential for a successful collision is proportional to the colloid concentration (the fraction of the colloids swept up is constant for a given number of collisions) e 2 Separate variables 1 2 3 N cviscous 4.8Floc t 1 N cinertial e 3 9 5.6Floc 2 t d Floc 8 Integrate from initial colloid concentration to current colloid concentration Classic first order reaction with number of successful collisions replacing time CColloid ln CColloid 0 Laminar Flow Cases kN c 1 2 2 e 3 N cviscous 4.8Floc t Only colloids can collide effectively (big flocs are useless) 1 2 e 3 N cviscous 4.8Floc t 2 2 3 Floc volume fraction that matters for successful collisions is the colloid (small floc) fraction 1 2 CColloid0 1 Colloid t ln 4.8k CColloid e C Colloid Colloids can attach to all flocs 2 3 N1cViscous CColloid0 d 4.8t Colloid d0 1 Colloid t 4.8k CColloid0 2 3 d0 d 2 3 DFractal 2 3 DFractal 3 3 Floc e 1 2 1 2 CColloid0 ln C e Colloid d Floc0 d0 3 DFractal Floc volume fraction is a function of floc size (d) (which is a function of the reaction progression and shear conditions in the reactor Comparison of Flocculation Hypotheses If colloids could aggregate with all of the flocs then colloids would aggregate VERY rapidly 100 Turbidity (NTU) colloids aggregate everything aggregates 10 1 2 3 0.1 10 100 1000 10000 1 2 CColloid0 1 Colloid t ln 4.8k CColloid e CColloid Time (s) Ten state standards 30 minute flocculation time. The graph is laminar flow case, full scale flocculators are turbulent flow. t 1 Colloid 4.8k CColloid0 2 3 d0 d 2 3 DFractal 3 1 2 CColloid0 ln e CColloid 100 NTU clay suspension, = 0.1 Show that pC* is proportional to time for “everything aggregates” model Coiled Tube Flocculation Residual Turbidity Analyzer Dr. Karen Swetland Dissertation research Two Phase Floc Model: PACl pC * 2 3log(e) 2 W Gt 03 N C1 Coag 3 2 VCapture 2 pC* 1.5 q=1200s, 5 NTU q=1200s, 15 NTU q=1200s, 50 NTU q=1200s, 150 NTU q=1200s, 500 NTU q=800s, 5 NTU q=800s, 15 NTU q=800s, 50 NTU q=800s, 150 NTU q=800s, 500 NTU Fitted Model 1 0.5 0 0.1 Initial floc growth N C1 1 10 *G*q*^2/3 100 Two Phase Floc Model: Alum pC * 2 3log(e) 2 W Gt 03 N C1 Coag 3 2 VCapture 2 pC* 1.5 q=1200s, 5 NTU q=1200s, 15 NTU q=1200s, 50 NTU q=1200s, 150 NTU q=1200s, 500 NTU q=800s, 5 NTU q=800s, 15 NTU q=800s, 50 NTU q=800s, 150 NTU q=800s, 500 NTU Fitted Model 1 0.5 0 0.1 Initial floc growth N C1 1 10 *G*q*^2/3 100 Solving for pC*: Lambert W or ProductLog Function (Laminar flow) 2 3 CColloid0 CColloid0 ln CColloid CColloid ln C * C 2 *3 C* e 2 3 0 Gt Coag VCapture 2 Coag 3 2 W Gt03 2 3 VCapture 2 Coag 3 Gt0 VCapture Use Wolframalpha to solve for C* W is the Lambert W function 2 Coag 3log( e ) 2 * 3 pC W Gt0 2 VCapture 3 Log is base 10 Laminar Flow Floc Model Coag 0.4 mm Sedimentation s velocity of ??? Empirical Collisions to make first big flocs = 0.4 2 3log( e ) 2 Coag * 3 pC W Gt 0 N C1 2 3 VCapture Floc volume fraction Lambert W Function VCapture mm Sedimentation tank 0.12 capture velocity s Fractional surface coverage of colloid by coagulant Flocculation time e Velocity gradient 1 2 Turbulent Flow Case N cInertial CColloid ln kN c CColloid 0 1 3 8 e 9 5.6Floc 2 t d Floc Only colloids can collide and attach effectively 8 9 N cInertial C e 5.6t Colloid 2 d Colloid Colloid 1 3 Why are almost all of these collisions in the inertial range? Colloids can attach to all flocs 8 9 1 2 3 CColloid0 1 Colloid dColloid t ln 5.6k CColloid e CColloid 8 9 NcInertial CColloid0 d 5.6t Colloid d0 1 Colloid t 5.6k CColloid0 8 9 d0 d Floc 2 3 DFractal 3 2 3 DFractal 3 e 2 d Floc 1 1 3 d 3 CColloid0 ln e C Colloid 2 Floc Floc d Floc0 d0 3 DFractal Turbulent Flow Flocculator: “Big Flocs are useless” hypothesis 8 9 1 2 3 CColloid0 1 Colloid dColloid t ln 5.6k CColloid e CColloid After the big flocs become non reactive, then the average separation distance between the remaining flocs is larger and thus the collisions between active particles is dominated by inertia. Ten state standards 30 minute flocculation time. Turbidity (NTU) 100 3 NTU 10 NTU 30 NTU 100 NTU 300 NTU 10 The predicted performance is quite similar given a wide range of influent turbidities The predictions seem reasonable 1 0.1 100 1000 time (s) 10000 = 0.1 for these plots e = 2.6 mW/kg Turbulent Flow Flocculator: Colloids can attach to all flocs hypothesis 1 Colloid t 5.6k CColloid0 8 9 d0 d Floc 2 3 DFractal 3 1 2 d Floc 3 CColloid0 ln e CColloid Turbidity (NTU) 100 3 NTU 10 NTU 30 NTU 100 NTU 300 NTU 10 100 1000 time (s) Ten state standards 30 minute flocculation time. The predicted performance is quite different from what we observe. High turbidity would be very easy to treat if this hypothesis were true! Flocculators would be tiny! 1 0.1 10 Here the separation distance between reactive flocs might be less than the Kolmogorov length scale 10000 For these plots = 0.1 e= 2.6 mW/kg pC* proportional to time extra Solving for pC*: Lambert W or ProductLog (Turbulent Flow) 8 9 CColloid0 CColloid0 ln CColloid CColloid ln C * 8 * 9 C C* e te 8 9 0 d 1 3 2 3 Colloid 1 3 Coag te 2 VCapture 3 dColloid 8 9 0 Coag VCapture 1 8 9 8 te 3 Coag W 09 2 8 9 VCapture 3 d Colloid C* CColloid CColloid0 C pC * log Colloid CColloid 0 Use Wolframalpha to solve for C* W is the Lambert W function 1 8 3 9 log(e) 8 9 te Coag pC * W 0 2 9 8 VCapture 3 dColloid Log is base 10 extra 8 9 log(e) 8 9 te Coag pC * W 0 2 9 8 VCapture 3 dColloid Lambert W Function 1 3 3 pC* 2 1 0 0.1 1 10 100 1 8 3 Coag 8 9 te 9 0 2 VCapture 3 dColloid 1000 Performance varies very little over wide range of inputs. Diminishing returns on investment Proposed Turbulent Flocculation Sedimentation Model (missing phase 1) Flocculation time Energy dissipation rate Sedimentation velocity of ??? Lambert W Function 1 8 9log(e) 8 9 te 3 Coag * pC W 0 2 9 8 VCapture 3 dColloid -log(fraction remaining) Sedimentation tank capture velocity Characteristic colloid size Initial floc volume fraction Fractional surface coverage of colloid by coagulant What does the plant operator control? ________ VCapture te What does the engineer control? __________ 0 What changes with the raw water? __________ 1 3 Turbulent Flocculation Sedimentation Model Questions Coag Must represent a characteristic sedimentation velocity of the floc suspension. It could be the average terminal velocity of the full size flocs in the flocculator effluent 2 3 dColloid Is a characteristic size of a colloid. Is a function of the characteristic size of the coagulant precipitate, geometry of the colloids, and loss of coagulant to reactor walls Fractal Flocculation Conclusions It is difficult to flocculate to a low residual turbidity because the time between effective collisions increases as the number of colloids and nonsettleable flocs decreases Colloids don’t attach to full size flocs Perhaps because the surface shear on the big floc is too high for a colloid to attach (surface shear increases with diameter) If we could routinely break up full size flocs perhaps we could speed the aggregation process Perhaps related to deformability? What is the model missing? The model doesn’t include any particle aggregation that occurs in the sedimentation tank. Residence time in the sed tank is long and energy dissipation rate is low. Flocculation in the sed tank (especially if there is a floc blanket) could be very important. Thus real world performance is likely better than model predictions. q Flocculator Collision Potential 1 3 te dt 0 The majority of the collisions in a turbulent hydraulic flocculator are in the inertial range and the collision potential is determined by flocculator residence time, q, and energy dissipation rate, e. The collision potential is given the symbol and has the dimension of m2/3 and this length scale will be a property of the reactor geometry The next set of notes provides guidance for designing the geometry of a flocculator to obtain a target collision potential Required Collision Potential* VCapture d Coag 2 3 Colloid 8 9 8 9 CColloid0 CColloid0 ln C C Colloid Colloid q 1 3 te dt 0 Turbidity (NTU) 100 Agalteca, Alauca, Marcala 2 were designed to have =100 m2/3 3 NTU 10 NTU 30 NTU 100 NTU 300 NTU 10 We are currently using =75 m2/3 1 0 20 40 60 80 100 ψ (m^2/3) *The model has not been validated for turbulent hydraulic flocculator. Thus this is only a rough estimate. Review 1 8 3 Coag 9 log(e) 8 9 te * pC W 0 2 9 8 VCapture 3 dColloid Why is it that doubling the residence time in a flocculator doesn’t double pC* for the flocculator? Why does increasing floc volume fraction decrease the time between collisions? Which terms in the model are determined by the flocculator design? extra Surface coverage () of clay by coagulant precipitate In our modeling work we didn’t cover how to calculate what fraction of the clay surfaces are coated with coagulant. The following slides are the equation heavy derivation of that coverage. We assume coagulant precipitate has some characteristic diameter when it sticks to the clay and that the clay can be represented as a cylinder. We also assume that the coagulant sticks to everything including reactor walls. The coagulant approaches the clay surface randomly and thus accumulates in a Poisson distribution. The random bombardment means that some coagulant is wasted in double coverage of previously covered clay. extra Floc Model Equations for AClay ATotal SAClayTotal SAClayTotal SAWall SAClayTotal 4 2 Tube Tube D L ClaySphere AClay ATotal SAClay VClay 1 SAWall 1 SAClayTotal DClay 3 2 3 DTube LTube ClaySphere 2 DClay Surface area of clay divided by total surface area of clay + reactor walls Clay SAClay 6 SAClay DClay 2 VClay DClay VClay 6 SAClayTotal SAWall DTube LTube Clay SAClay VClay 6 ClaySphere DClay Surface area to volume ratio for clay normalized by surface area to volume ratio for a sphere extra Ratio of clay surface area to total surface area (including reactor walls) AClay ATotal 1 DTube LTube 1 2 3 DTube LTube ClaySphere 2 DClay AClay ATotal 1 1 2 DClay 3DTube ClaySphereClay Clay extra Poisson distribution and coagulant loss to walls combined to get surface coverage N CoagPerClay SAClay CCoag 1 VClay Clay Coag D 3 SAClay CClay Coag AClay ATotal 6 2 Coag D N CoagPerClay SAClay Coag DClay Clay DCoag ClaySphere 1 1 2 DClay 3DTube ClaySphereClay SAClay VClay 6 ClaySphere DClay The surface coverage is reduced due to stacking (which is handled by the Poisson distribution) and by the loss of coagulant to the reactor walls. 1 e 2 DCoag NCoagPerClay SAClay AClay ATotal loss of coagulant to the reactor walls 1 e Coag DClay AClay ATotal Clay DCoag ClaySphere extra Clay platelet geometry HD VPlatelet VPlatelet H ClayPlatelet DClayPlatelet 4 6 2 ClayPlatelet D 4 VPlatelet HD H ClayPlatelet 4 3 DClayPlatelet D.Clay is the diameter of a sphere with equal volume as the clay platelet 3 DClay VPlatelet HD DClayPlatelet Model clay as cylinder with height and diameter 3 ClayPlatelet D 1 3 6 3 DClay 2 D.ClayPlatelet is the diameter of a DClay cylinder given ratio of height to diameter 3 HD and equal volume spherical diameter extra Clay platelet geometry 1 3 SAClayPlatelet 2 SAClayPlatelet 4 DClayPlatelet 2 DClayPlatelet DClayPlatelet H ClayPlatelet 2 3 2 3 2 DClay 3 HD HD 2 2 2 2 2 DClay HD DClay 4 3 HD 3 HD 2 3 SAClayPlatelet 1 2 2 HD D Clay 2 3 HD ClaySphere SAClayPlatelet D 2 Clay ClaySphere 1 2 HD 2 3 HD 2 3 H ClayPlatelet DClayPlatelet extra Surface coverage of clay 1 e 1 e Coag DClay AClay ATotal Clay DCoag ClaySphere Coag DClay 1 Clay DCoag AClay ATotal 1 ClaySphere 1 e Coag DClay 1 Clay DCoag 1 e 3 DTubeClay 1 2 3 1 2 HD 2 3 HD 3 DTubeClay Coag DClay 1 Clay DCoag 3DTube ClaySphereClay 2 DClay ClaySphere 1 1 2 DClay 2 DClay 1 2 HD 2 3 HD AClay ATotal ClaySphere 1 ClaySphere 2 DClay 3DTubeClay 1 2 1 2 3 HD 2 3 HD 2 3 If you neglect wall loses extra Solving for Coagulant Dose 1 e CCoag Coag DClay 1 Clay DCoag 1 2 3 2 DClay 1 2 HD 2 3 HD 3 DTubeClay Coagulant dose is inside coagulant volume fraction Solve for dose… 2 3 2 DClay DCoagClay 1 2 Coag HD ln 1 2 3 HD 3DTubeClay DClay d 2 3 Colloid 8 9 8 9 VCapture CColloid0 CColloid0 ln Coag CColloid CColloid Given a target residual turbidity, solve for required , then solve for coagulant dose extra Fraction of coagulant on clay Loss of clay to reactor walls 1 0.1 5 NTU 50 NTU 500 NTU 5000 NTU 0.01 0.001 0.001 0.01 0.1 1 10 Tube diameter (m) Loss of coagulant to reactor walls can dominate for low turbidities and small reactors. For the LFSRSF in India treating 5 NTU water and injecting coagulant into 7.5 cm pipes the clay only gets 23% of the applied coagulant! There may be a way to design a larger contact chamber for the coagulant to reduce losses to the walls. extra Effect of stacking and wall loss in a 10 cm diameter flocculator with 20 NTU Fractional clay surface coverage 1 0.8 0.6 0.4 0.2 0 0 1 2 3 4 5 Coagulant concentration (mg/L) No Stacking and No Wall Loss No Wall Loss Stacking and Wall Losses Stacking due to random bombardment of the clay surface with coagulant nanoglobs. Stacking becomes significant for high surface coverage. Wall losses also cause a significant reduction in clay surface coverage.