Presentation on theme: "Lecture 6: Water & Wastewater Treatment"— Presentation transcript:
1 Lecture 6: Water & Wastewater Treatment Objectives:Define primary, secondary, and tertiary treatmentDefine BODDescribe the activated sludge processSetup and solve a mass balance for an activated sludge system
2 Review Sorption: Settling Kd=Cs/CL CT=(1+KdCss)CL Fraction sorbed vs. fraction remaining in waterSettlingSettling velocity:Percent of particles removed: (1-Css/Css,o) x 100%Where,
3 Well-Mixed Settling Tank vsCssQ, Css,oQ, CssVSuspended solids remaining:Define the Overflow Rate:~ 20 – 100 m/day in treatment plants
4 Wastewater Treatment POTW – Publicly Owned Treatment Works 0.4 – 0.6 m3/person/day15 million people in Los Angeles 7.5 x 106m3/day or 2000 MGD (million gallons per day)Hyperion – 450 MGDClean Water Act (CWA) – 1977 – Set effluent (what is released by treatment plants into the environment) standards
5 Stages of Water Treatment PrimaryContaminants (60% of solids and 35% of BOD removed)Oil & GreaseTotal Suspended Solids (Css or TSS) – 60% RemovedPathogensBOD – 35% removedProcessesScreensGrit SettlingScum FlotationPrimary Settling
6 Stages (continued) Secondary Contaminants Processes BOD – 90% Removed TSS – 90% RemovedProcessesTrickling Filter – rotating diskActivated Sludge – Suspended and mixedOxidation ponds – lagoons(promote contact between microbes and contaminants)
10 Primary Sludge (cont’d) Q, Css,oQ, CssGiven:Q = 4000 m3/dCss,o = 200 mg/L and Css = 100 mg/LSludge density = 0.05 kg/LOverflow rate of 50 m/dFindPopulation of town served by this unitSludge production rateArea of settling tankSettling velocity of particlesCut-off size of particles (find the particle diameter corresponding to this settling velocity. Assume rs = 2600 kg/m3. All particles larger than this size will settle)
15 Definition of BODMicroorganisms (e.g., bacteria) are responsible for decomposing organic waste. When organic matter such as dead plants, leaves, grass clippings, manure, sewage, or even food waste is present in a water supply, the bacteria will begin the process of breaking down this waste. When this happens, much of the available dissolved oxygen is consumed by aerobic bacteria, robbing other aquatic organisms of the oxygen they need to live. Biological Oxygen Demand (BOD) is a measure of the oxygen used by microorganisms to decompose this waste. If there is a large quantity of organic waste in the water supply, there will also be a lot of bacteria present working to decompose this waste. In this case, the demand for oxygen will be high (due to all the bacteria) so the BOD level will be high. As the waste is consumed or dispersed through the water, BOD levels will begin to decline.
16 Activated Sludge Nomenclature Q+QR, S, XQ-Qs, SQ, So, XoQs+QR, XsQR, XsQs, XsS stands for conc. of substrate (organic matter, waste, etc.) or BODX stands for conc. of microorganisms
17 Activated Sludge Nomenclature (cont’d) Q, So, XoQ+QR, S, X~Q, Sm, VQs+QR, XsQR, XsQs, XsAssumptions:Effluent bacteria concentration is 0Concentration of substrate or BOD in sludge is 0Sludge flowrate (Qs) is much smaller than Q
18 Decay of BOD and growth of organisms Substrate or BOD (S) decays with rate k:Microbes (X) grow at rate m:
19 Activated Sludge Equations The following equations are derived from conducting mass balances over:The entire systemThe aeration tankThe sedimentation tankAny good book on wastewater engineering will have the derivations if you are curious!
20 Activated Sludge Equations Biomass (X) balance over entire system:Substrate (S) balance over entire system:
21 More AS equations Mass balance over sedimentation tank: Other equation(s)/rules of thumb:F/M = QSo/XV - Food-to-microbe ratio: 0.3 – 0.7 d-1QR ~ 0.25 – 0.50 x QX ~ 1000 – 2000 mg/LProblem types:Given Q, So, and S (target concentration)Find QR, Qs, X, m, V, Y
22 Example Find Qs, m, V, Y Given: Q = 1000 m3/d So = 150 mg/L S = 15 mg/LQR = 240 m3/dF/M = 0.3 d-1X = 2000 mg/LXs = 1% or 10,000 mg/L