Wastewater sedimentation

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Wastewater sedimentation

  1. 1. SEDIMENTATIONSettling phenomena involved in an urban wastewater treatment plant (UWWTP) Lectures for the course of “Wastewater Treatment”Second Cycle Degree (MSc Level) in Environmental Engineering University of Padua, ITALY Prof Alessandro SPAGNI 29/11/2012 04/12/2012 Padua Sabino DE GISI ENEA
  2. 2. Framework• Introduction• Type of sedimentation processes• Primary and secondary sedimentation in an urban wastewater treatment plant (UWWTP)• Goal of lessons• Solid-Flux Analysis• Technologyes of sedimentation tanks• Exercises • Design of a primary sedimentation unit • Design of a secondary sedimentation unit in a CAS system (convenctional activated sludge)
  3. 3. IntroductionWhat is sedimentation?• Sedimentation is the separation of suspended particles from water by gravitational settling;• The primary purpose is to produce a clarified effluent;• It is one of the most widely used unit operations in wastewater treatment. It is used for: • grit removal, particulate-matter removal in the primary settling basin; • biological floc-removal in the activated sludge settling basin; • solids concentration in sludge thickeners.
  4. 4. Type of sedimentation Settling phenomena involved in wastewater treatment• According literature (Metcalf & Eddy, 2003), 4 type of settling can occur: Clear water region • discrete particle; Discrete settling region (type 1) • flocculant; Depth (h) Discrete settling region (type 2) • hindered (also called zone); Hindered zone • compression. Compression• For urban wastewater, the region attention is focused above all Time (t) discrete particle sedimentation and hindered/flocculant particle sedimentation.
  5. 5. Type of sedimentation Settling phenomena involved in wastewater treatment (Metcalf & Eddy, 2003)
  6. 6. Type of sedimentationSettling phenomena involved inwastewater treatment Water processing flow diagram for a large UWWTP
  7. 7. Type of sedimentationSettling phenomena involved inwastewater treatment Sludge processing flow diagram for a large UWWTP
  8. 8. Primary and Secondary sedimentation Primary sludge characterization Primary sludge Parameter Unit Range Most frequent value Production g/ab/year 30 - 90 50 Suspendid solids kgSS/m3 30 - 120 50 Volatile solids %SS 65 - 90 75 - 80 Calorific power kCal/kgSS 3,800 – 5,600 4,350 N %SS 1,5 - 5 2,5 P as P2O5 %SS 0,5 – 2,8 1,6 Secondary biological sludge characterization Secondary sludge Parameter Unit Range Most frequent value Production g/ab/year 30 - 50 40 Suspendid solids kgSS/m3 5 - 20 15 Volatile solids %SS 55 - 90 80 Calorific power kCal/kgSS 2,700 – 4,500 3,600 N %SS 5 - 10 7-8 P as P2O5 %SS 3 - 11 7 Metcalf & Eddy (2003)
  9. 9. Goal• The goal of this lesson is the design of the secondary sedimentation tanks and the presentation of the Solids-Flux theory;• The design of primary sedimentation tanks will be developed in the next lession with the use of exercises.
  10. 10. Solid-Flux AnalysisSome information• Is a method for calculation the area required for hindered settling based on an analysis of the solids (mass) flux;• Data derived from settling tests must be available when applying this method;• Work hypthotesis is a settling basin operating at steady state with a constant flux of solids in moving downward;• The moviment of solids is due to 2 contribution: • gravity (hindered) settling; • bulk transport due to the underflow being pumped out and recycled.
  11. 11. Solid-Flux Analysis Symbols Influent Effluentq = inlet flowrateqf = returnactivated sludge Clarified WW BorderlineX0 = SSconcentration in theoxidation basin SludgeXf = SS Section (i) withconcentration of surface Aactivated sludge Z Activated SludgeWith reference a generic section (i) with a fixed value of the concentration xi,solid-flux (SF)i is defined as the quantity of solids that crosses an horizontalsurface unit per unit of time:SF = SF1 + SF2 = (xi · vi) + (xi · qf/A) = (xi · vi) + (xi ·u) with u = qf/A = costwhere:SF1 = solid flux due to gravity – mass sedimentation (M L-2 T-1)SF2 = solid flux due to the extraction of the sludge from the bottom of the tank (M L-2 T-1)
  12. 12. Solid-Flux Analysis x1 < x2 < x3 < xi Calculation of SF1 (mass sedimentation solid flux) h(t1) • The determination of SF1 is carried out with a series of laboratory cylinders in which mixed aeration samples h(t2) are introduced with different values of solids h(t3) concentration (x). • For each cylinder, the position of the interface water/sludge is reported, as a function of time. h(t4) • As visible in figure 1, mass sedimentation velocity for the initial concentration of the slurry (xi), is calculated as is the angular coefficient of the straight lineh v SF1 SF1(i) = xi · vi v3 v2 v1 v1 x3 v2 x2 x1 t x1 x2 x x Fig. 1 Fig. 2 Fig. 3
  13. 13. Solid-Flux Analysis Calculation of SF2 (extraction solid flux) SF2 SF2 = xi · qf/A = xi ·u with u = qf/A = cost where SF2 is the equation of a straight line passing SF2(i) through the origin and with the angular coefficient equal to u Calculation of SF (solid flux) x SF = SF1 + SF2 First of all, the total solid-flux curve has a maximum value. Then, a minimum value can be observe.SF This minimum value is called limiting flux (SFL). SF The sedimentation tank must be fed with a flow value less than the limiting flux (SFL). Otherwise, the solids exit out of the tank within the clarified effluent.SFL The tank surface useful for a correct function of secondary sedimentation (A) is equal to: SF2 (q+qf) · x0 A= SF1 SFL xL xf x
  14. 14. Solid-Flux Analysis Determination of Return Activated Sludge flowrate (qf) - RAS Influent x0 q + qf Effluent Oxidation basin Secondary (CAS System) Air sedimentation tank qf, xf RAS FluxMass balance on V.C. (steady state condition) , xf to sludge line MSS,In +/- Gen = MSS,Out + ∆(t) Ricircolation 0 0 RAS ratio x 0 = MSS,In – MSS,out 0 = x0 ⋅ q0 + qf ⋅ xf – (q+qf) ⋅ x qf = xf - x ·q 0
  15. 15. Solid-Flux Analysis How we can use the Solid-Flux Analysis for the design of a new wastewater treatment plant? Influent x0 q + qf Effluent Oxidation basin A Air Secondary (CAS System) sedimentation tankGoal qf, xfDesign of surface (A) of the RASsecondary sedimentation tank ina CAS system.HypotesisSludge data (speed, concentration) regardingthe project wastewater (and the mixed liquor)are taken from literature SF = SF1 + SF2 = (xi · vi) + (xi · qf/A) = (xi · vi) + (xi ·u)
  16. 16. Solid-Flux AnalysisDesign of the secondary sedimentation tank with Solid-Flux Analysis1. Data input (activated sludge at different value of xi) x (kgSS/m3) 1 1.5 2 3 4 5 6 8 10 v (m/h) 6.72 6.10 4.80 2.40 1.00 0.55 0.34 0.15 0.07 FS1 (kgSS/m2/h) 6.72 9.15 9.60 7.20 4.00 2.75 2.04 1.20 0.70 2. Interpolation curve (vi ; xi) 3. SF1 curve q = inlet flowrate to the CAS system (m3/s) Solid-Flux SF (kgSS/m2/h) x0 = oxidation basin concentration (kgSS/m3) Experimental Speed vi (m/h) curve SF1 i.e. 4-6 kgSS/m3. xf = target = value of RAS concentration (kgSS/m3) i.e. 8-12 kgSS/m3. Concentration xi (kgSS/m3) Concentration xi (kgSS/m3)
  17. 17. Solid-Flux AnalysisDesign of the secondary sedimentation tank with Solid-Flux Analysis4. SFL calculation 5. SF2 and u calculation Solid-Flux SF (kgSS/m2/h) SFL SF2 SFL Q SFL P u xf Concentration xi (kgSS/m3) xf xf SFL U= xf
  18. 18. Solid-Flux AnalysisDesign of the secondary sedimentation tank with Solid-Flux Analysis6. SF calculation 7. Calculation of qf Solid-Flux SF (kgSS/m2/h) SF1 SF x qf = ·q SFL xf - x SF2 8. Calculation Surface (A) xf (q+qf) · x0 A= Concentration xi (kgSS/m3) SFL The value of A surface of the secondary sedimentation tank allows to thicken the sludge to the xf value fixed
  19. 19. Technology Large urban wastewater treatment plantSecondary treatment Primary treatment
  20. 20. TechnologyOxidation basins WW distribution well Primary sedimentation tanks
  21. 21. Technology Primary sedimentation tanks
  22. 22. Technology Thomson effluent weir 150 50 150 50 90° 75 150 (values in cm)
  23. 23. TechnologySecondary sedimentation tanks
  24. 24. Technology Secondary sedimentation tanks
  25. 25. Technology Bridge Drive unit Surface skimmer Sludge scrapers Inlet pipe Secondary sedimentation tanks
  26. 26. Technology Effluent weir Support Scum baffle Secondary sedimentation tanks
  27. 27. Technology Scum box Inlet scum Scum pipe Versus Scum pitSecondary sedimentation tanks
  28. 28. Technology Rotation of bridge Scum box Route of Scum Secondary sedimentation tanks
  29. 29. Technology Secondary sedimentation tanks Clarified wastewater’s route Surface skimmer Effluent weir Scum baffleEffluent launder
  30. 30. TechnologyCockpitdivider Secondary sedimentation tanks
  31. 31. Technology Cockpit divider Secondary sedimentation tanks
  32. 32. Technology Thomson effluent weir Central pivot Bridge Scum boxDistribution system Secondary sedimentation tanks
  33. 33. Exercise 1Design of primary sedimentation unitsDesign the primary sedimentation units of a large wastewater treatment plant serving65,000 ab. In order to ensure continuity of operation, two equal size units should berealized. In particular, calculate: the geometry of the single sedimentation tank.1. Some consideration Desing Flowrateq q PM q PM q PB q PM Trattamenti biologici Trattamenti preliminari Trattamenti primari Disinfezione e terziari q PM - qPB q - q PM q PM qDesign flowrates considered in a wastewater treatment plant Ricettore qPM = max flowrate inlet in the plant (q24)C = average flowrate inlet in the plant
  34. 34. Exercise 1Design of primary sedimentation units1. Some consideration Characteristics parameters Range on Range on Parameter Unit (q24)C qPM Primary sedimentation followed by biological secondary treatment HRT (τ) h 2–3 0,66 – 0,83 3 2 Surface hydraulic load (Cis) m /m /h 1.25 – 2.08 3-5 Depth(h) m 2-5 - 3 Weir load (Cs) m /m/d 125 - 500 - Characteristics parameters for the design and verification of a primary sedimentation tank (Metcalf & Eddy, 2003)
  35. 35. Exercise 1Design of primary sedimentation units2. Data input and design parametersThe following parameters are considered: equivalent population = 65,000 PE; max flowrate inlet in the plant (qPM) = 65,000 m3/d; average flowrate (q24)C = 13,000 m3/d; number of tanks (N) = 2; shape of single tank: radial; Cis,max (on qPM) = 5 m3/m2/h; τmax (on qPM) = 2 h; hmin = 2.5 m; CS = 125-500 m3/m/d.3. Calculation of the minimum sedimentation tanksurface(Ssed,min)The following calculation are devepoled with refer to a single unit and considering theseflowrate values: qPM/2 = 32,500 m3/d = 1,354.16 m3/h; (q24)C/2 = 6,500 m3/d = 270.83 m3/h.
  36. 36. Exercise 1Design of primary sedimentation units4. Calculation of the minimum sedimentation tanksurface(Ssed,min)The minimum sedimentation tank surface is equal to: q PM 1,354.16(m 3 /h) S sed, min = = = 270.83 m 2 C is, max ⋅2 5(m 3 /m 2 /h)5. Calculation of the real diameter (Dreal) and the real surface(Sreal) of the single sedimentation tankThe real diameter (Dreal) of the single tank is calculated from the minimum diameter(Dsed,min). Its value is equal to: 4 ⋅ S sed,min 4 ⋅ 270.83 D sed,min = = (m) = 18.57 m → 19 m π 3.14 π ⋅ D2 3.14 ⋅ 19 2 (m 2 ) S real = real = = 283.38 m 2 4 4
  37. 37. Exercise 1Design of primary sedimentation units6. Verification of Surface Hydraulic Load (Cis) with reference(q24)CWith reference (q24)C/2, the Surface Hydraulic Load (Cis) for the single tank is less thanthe maximum allowed value: (q 24 ) C 270.83(m 3 /h) C is = = = 0.955 m 3 /m 2 /h < 2 m 3 /m 2 /h S real ⋅ 2 283.38(m 2 )7. Calculation of the volume and depth of the single tank andchoice of the commercial tankFor volume calculation, a hydraulic detention time (HRT) of 40 min (0.66 h) withreference qPM/2 flowrate is considered. The single tank volume is equal to: q PM V= ⋅ τ max = 1,354.16 (m 3 /h) ⋅ 0.66 (h) = 893.75 m 3 2 V 893.75(m 3 ) h= = 2 = 3.15 m > 2.5 m S real 283.38(m )
  38. 38. Exercise 1Design of primary sedimentation units7. Calculation of the volume and depth of the single tank andchoice of the commercial tankOnce determined values of the Table 1. Technical data of radial sedimentation unit “type PRTP” of Ecoplants Inc.principal geometrical variables(volume, diameter and surface), Diameter and depth [m] Engine Flowrate Model Surface [m2] 3 power [m /d] D hchoice of the commercial settler [kW] PRTP-50 19.6 480 5 3.6 0.12model should be made. With PRTP-60 28.3 690 6 3.6 0.12reference to the real diameter of PRTP-70 38.5 940 7 3.6 0.12 PRTP-80 50.3 1,230 8 3.6 0.1219m and considering table 1, PRTP-90 63.6 1,550 9 3.6 0.18Ecoplants PRTP-190 tank is PRTP-100 78.5 2,240 10 3.5 0.18 PRTP-110 95.0 2,710 11 3.5 0.18assumed. PRTP-120 113.1 3,220 12 3.5 0.18The final characteristic parameters PRTP-130 132.7 3,780 13 3.5 0.25 PRTP-140 153.9 4,390 14 3.5 0.25of the single sedimentation tanks PRTP-150 176.7 5,040 15 3.5 0.25are: PRTP-160 201.1 5,730 16 3.5 0.25 PRTP-170 227.0 7,400 17 3.2 0.25 h = 3.2 m; PRTP-180 254.5 8,300 18 3.2 0.37 D = 19 m; PRTP-190 283.5 9,240 19 3.2 0.37 PRTP-200 314.2 10,240 20 3.2 0.37 S = 283.5 m2; PRTP-210 346.4 11,290 21 3.2 0.37 V = 283.5 (m2) x 3.2 (m) = PRTP-220 PRTP-230 380.1 415.5 12,390 13,540 22 23 3.2 3.2 0.37 0.37 907.2 m3; PRTP-240 452.4 14,750 24 3.2 0.55 PRTP-250 490.9 16,000 25 3.2 0.55 Engine power = 0.37 kW. PRTP-260 530.9 17,310 26 3.2 0.55 PRTP-270 572.6 18,660 27 3.2 0.55 PRTP-280 615.8 20,070 28 3.2 0.55
  39. 39. Exercise 1Design of primary sedimentation units7. Calculation of the volume and depth of the single tank andchoice of the commercial tank View of “type PRTP-190” sedimentation tank (Ecoplants Inc., Italy)
  40. 40. Exercise 1Design of primary sedimentation units8. Verification of HRT on (q24)C V 907.2(m3 ) τ min = = = 3.35 h (q 24 )C /2 270.83(m /h) 3The obtained value is greater than the maximun value generally considered on (q24)C =3h. With this solution a more safety margin is guaranteed above all during the peakperiod (qPM).9. Verification of weir load on (q24)CThe last verification regarding the weir load (CS): (q 24 )C /2 6,500(m3 /d) Cs = = = 108.9 m 3 /m/d < 500 m 3 /m/d Ls 59.7(m)where LS is the length of the tank circumference: LS = D ⋅ p = 19 (m) ⋅ 3.14 = 59.7 m
  41. 41. Exercise 1Design of primary sedimentation units10. Calculation of primary sludge production (Psludge)The following parametersare considered: average flowrate (q24)C = 13,000 m3/d; %TSS = 65% Inlet TSS Percentage removal [%] concentration (TSS,in) = 350 gTSS/m3; Percentage TSS removal (ηTSS) = 65%; Primary sludge solids (S) = 4%. τ = 3,35 h Time [h] (q24)C
  42. 42. Exercise 1Design of primary sedimentation units10. Calculation of primary sludge production (Psludge) (q 24)C Influent TSSin M TSS, in (q 24)C(q24)C = 541,66 m3/h TSSin = 350 g/m3 TSS out Effluent PTSS,in = ? M TSS, out (q24)C = 541,66 m3/h TSSout = ? g/m3 PTSS,out = ? M TSS, rimossi (secco) MFanghi (secco + umido) Primary sludge PTSS,removed = ? (only dry matter) Psludge = ? (dry matter + water)
  43. 43. Exercise 1Design of primary sedimentation units10. Calculation of primary sludge production (Psludge) (q 24)C Influent TSS in M TSS, in (q 24)C(q24)C = 541,66 m3/h TSS out Effluent TSSin = 350 g/m3 M TSS, out (q24)C = 541,66 m3/h PTSS,in = ? TSSout = ? g/m3 PTSS,out = ? M TSS, rimossi (secco) MFanghi (secco + umido) Primary sludge 1) PTSS, in = (q24)C ⋅ TSSin = 541,66 (m3/h) ⋅ 350 (g/m3) = 189,6 kg/h 2) PTSS,out = MTSS, in ⋅ (1- ηTSS) = 189,6 (kg/h) ⋅ (1- 0,65) = 66,3 kg/h 3) PTSS,removed = MTSS, in - MTSS, out = (189,6 – 66,3) kg/h = 123,2 kg/h
  44. 44. Exercise 1 Design of primary sedimentation units 10. Calculation of primary sludge production (Psludge) (q 24)C Influent TSS in M TSS, in (q 24)C (q24)C = 541,66 m3/h TSS out Effluent TSSin = 350 g/m3 M TSS, out (q24)C = 541,66 m3/h PTSS,in = ? TSSout = ? g/m3 PTSS,out = ? This is the solid part of primary M TSS, rimossi (secco) sludge MFanghi (secco + umido) Primary sludge1) PTSS, in = (q24)C ⋅ TSSin = 541.66 (m3/h) ⋅ 350 (g/m3) = 189.6 kg/h2) PTSS,out = MTSS, in ⋅ (1- ηTSS) = 189.6 (kg/h) ⋅ (1- 0.65) = 66.3 kg/h Water?3) PTSS,removed = MTSS, in - MTSS, out = (189.6 – 66.3) kg/h = 123.2 kg/h + Solids ?
  45. 45. Exercise 1Design of primary sedimentation units10. Calculation of primary sludge production (Psludge) Water? PTSS PTSS S= = Psludge PTSS + Pwater Solids ? Psludge = PTSS,removed/S = 123.2 (kg/h)/0.04 = 73,937.5 kgsludge/dFor the calculation of qsludge (volumetric flowrate of primary sludge), the sludgedensity (ρ)must be considered. Generally, its value is about 1,000 kg/m3. Thishyphotesis is true if the wastewater solids value (S) is less than 10%. qsludge = Psludge/ρsludge = 73,937.5 (kg/d)/ 1,000 (kg/m3) ≅ 74 m3/d
  46. 46. Exercise 2 Design of secondary sedimentation units with the Solid-Flux Analysis1. Data input (activated sludge at different value of xi) x (kgSS/m3) 1 1.5 2 3 4 5 6 8 10 v (m/h) 6.72 6.10 4.80 2.40 1.00 0.55 0.34 0.15 0.07 FS1 (kgSS/m2/h) 6.72 9.15 9.60 7.20 4.00 2.75 2.04 1.20 0.70 2. Interpolation curve (vi ; xi) 3. SF1 curve q = 125 (l/s) x0 = oxidation basin Solid-Flux SF (kgSS/m2/h) concentration = 3.4 kgSS/m3 Experimental Speed vi (m/h) xf = RAS concentration curve SF1 = 12 kgSS/m3 Concentration xi (kgSS/m3) Concentration xi (kgSS/m3)
  47. 47. Exercise 2 Design of secondary sedimentation units with the Solid-Flux Analysis4. SFL calculation 5. SF2 and u calculation Solid-Flux SF (kgSS/m2/h) SFL = 3.6 SF2 kgSS/m2/h Q SFL P u xf Concentration xi (kgSS/m3) SFL 3.6 kgSS/m2/h xf = 12 kgSS/m3 xf U= = = 0.300 m/h 12 kgSS/m3
  48. 48. Exercise 2 Design of secondary sedimentation units with the Solid-Flux Analysis6. SF calculation 7. Calculation of qf Solid-Flux SF (kgSS/m2/h) SF1 SF SFL x 3.4 kgSS/m3 qf = ·q = · 125 (l/s) = 49 l/s SF2 xf - x (12-4) kgSS/m3 xf Concentration xi (kgSS/m3) 8. Calculation Surface (A) (q+qf) · x0 2,130 A= = = 592 m2 SFL 3.6
  49. 49. References• Bonomo L., Trattamenti delle acque reflue, McGraw-Hill Companies, Srl, Publishing Group Italia, Milano, ISBN: 978-88-386-6518-9, 2008 (in Italian).• Masotti L., Depurazione delle acque, tecniche ed impianti per il trattamento delle acque di rifiuto, 2nd Edizione, Edizioni Calderini, Bologna, Italia, ISBN: 88-7019-292-X, 1993 (in Italian).• Metcalf & Eddy, Wastewater Engineering. Treatment and Reuse, 4th ed., McGraw Hill, New York (USA), 2003.• IRSA-CNR, Istituto di Ricerca Sulle Acque - Consiglio Nazionale delle Ricerche, La protezione delle acqua dagli inquinamenti - Quaderno 2 - Aspetti biochimici e microbiologici dei processi depurativi naturali ed artificiali delle acque di rifiuto, Roma, 1974 (in Italian).• Passino R., La conduzione degli impianti di depurazione delle acque di scarico, Ed Scient. A Cremonese, Roma, 1980 (in Italian).• De Feo G., De Gisi S., Galasso M. (2012), Acque reflue, Progettazione e gestione di impianti per il trattamento e lo smaltimento, Dario Flaccovio Editore Srl, ISBN 9788857901183, 1244 pagine (in Italian). (http://www.darioflaccovio.it/libro.php/acque-reflue-df0118_C762)
  50. 50. Sabino DE GISI, Ph.D.ENEA(Italian National Agency for the New Technology, Energy and Sustainable EconomicDevelopment), Technical Unit on Models, Methods and Technologies for theEnvironmental Assessment (UTVALAMB), Water Resource Management DivisionVia Martiri di Monte Sole 4, 40129 Bologna, ITALYsabino.degisi@enea.it

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