1 / 44

BIOCHEMICAL OXYGEN DEMAND

BIOCHEMICAL OXYGEN DEMAND. BIOCHEMICAL OXYGEN DEMAND (BOD). BOD the amount of oxygen required by bacteria while stabilizing decomposable organic matter under aerobic condition. Organic matter serve as food for bacteria. Organic matter. Energy +. oxygen.

dwinslow
Download Presentation

BIOCHEMICAL OXYGEN DEMAND

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. BIOCHEMICAL OXYGEN DEMAND D.Akgul

  2. BIOCHEMICAL OXYGEN DEMAND (BOD) • BOD the amount of oxygen required by bacteria while stabilizing decomposable organic matter under aerobic condition. Organic matter serve as food for bacteria Organicmatter Energy + oxygen D.Akgul

  3. BOD testusedtodeterminepollutionalstrenght of domesticandindustrialwastes in terms of oxygenthattheywillrequireifdischargedintonaturalwatercources wastewater DO O2 Org. matter D.Akgul

  4. BOD test • Bioassayprocedure. • Measuresoxygenconsumedbylivingorganismswhileutilizingorganicmatterpresent in waste. • Conditions  as similar as possibletothosethatoccur in nature • To be quantitative protectfromtheairtopreventreaeration. • Env. conditionsshould be suitableform.o. • No toxicsubstance. D.Akgul

  5. D.O should be presentthroughoutthe test. O2solubility~9 mg/L DO used ~ amount of organic in sample Strongwastesmust be diluted D.Akgul

  6. Allnecessarynutrientsneededforbacterialgrowth. (N,P, traceelements) • Diversegroup of m.org. (seed) carrytheoxidationto CO2 • BOD test  a wetoxidationprocess • Livingorganismsserve as themediumforoxidation of org. mattertowaterandCO2 D.Akgul

  7. Oxygenrequirementtoconvert org. compoundto CO2 , water , ammonia. CnHaObNc + ( n + a/4 - b/2 – 3c/4 ) O2 nCO2+ (a/2 – 3c/2)H2O+ cNH3 • Temperatureeffectsareheldconstantbyperformingthe test at 20C D.Akgul

  8. İnfinite time is requiredforcompletebiologicaloxidation of organicmatter • Practicallythereactionconsideredcomplete in 20 days =>BOD ultimate • Largepercentage of the total BOD is exerted in 5days • The test has beendeveloped on thebasis of a 5-dayincubationperiod.=> 5 day BOD • 5 day test wasselectedalsoto minimize interferencefromoxidation of ammonia. D.Akgul

  9. BOD ReactionKinetics • Firstorderrxnkinetics Rxn rate proportionaltotheamount of oxidizableorganicmatterremaining at any time. (Ifmopopulation is nearlyconstant) • -dc /dt ˜ C • -dc/dt = k’C C: Conc. of oxidizable org matter K: rate constant • Rate decreases as C decreases D.Akgul

  10. In BOD considerations; • C L (ultimatedemand) • -dL/ dt=k’L • Lt/L = e-k’t • Lt=L* e-k’t • e-k’t= 10–kt • k=k’/ 2.303 D.Akgul

  11. k’ : BOD rxnconstant (d-1). Typically k’ ˜0.1 to 1 d-1 fordegradable org. matter in naturalwater at ambienttemp 10 < T < 30 °C BOD t = L (1-10–kt ) BOD t : BOD at any time t, L: BOD ultimate D.Akgul

  12. Theoxygenconsumed in the time intervalfromzeroto t is called BOD t • BOD t : BOD consumedwithin time t • Lt : Potential BOD remaining at time t • BODt = L –Lt = L – L. e-k’t = L (1- e-k’t) D.Akgul

  13. Example 1: Determine 1 day BOD andultimate BOD for a wastewaterwhose 5 day , 20 °C BOD is 200 mg/ L. • Thereactionconstant k ( base e) = 0.23 d-1 D.Akgul

  14. Itspossibletodeterminereactionconstant k at a temp. otherthan 20 °C. • VantHoff – Arrheniusrelationship : • Kt = k20 Ɵ ( t-20) • Ɵ = 1,056 (20<T<30 °C) • Ɵ = 1,135 ( 4<T<20 °C) D.Akgul

  15. Example 2: Calculate rate constantfor T=24C k=0.23 d-1 (@ 20C) Ɵ=1.047 D.Akgul

  16. Oxygenused Organicmatteroxidized ( DirectRatio) ~ Oxygenused D.Akgul

  17. Influence of Nitrification on BOD • BOD oroxygen-usedcurve is similartoorganicmatteroxidationcurveduringthefirst 8-10 days. • Forpropermeasurement of BOD, culturesused in BOD test shouldcontainheterotrophicbacteriaandotherorganismsthatutilizethecarbonaceousmatter. D.Akgul

  18. Additionallytheycontaincertainautotrophicbacteria, particularlynitrifyingbacteria. • nitrifyingbacteriaoxidizesnoncarbonaceousmatterforenergy • Nitrifyingbacteriapresentsmallamounts in untreateddomesticww. • @20 °C theirpopulations do not becomesufficientlylargetoexertoxygendemanduntil 8-10 days. • Interference of nitrificationeliminatedbytaking test period 5 days  BOD 5 D.Akgul

  19. D.Akgul

  20. Untreateddomesticwwcontainslittleamount of nitrifyingbacteria, but effluentfrombiological treatment unitssuch as ActivatedsludgeandTricklingfilter do containnitrifyingbacteriathat can alsoconsume D.O. in 5 days. • Inhibittheaction of nitrifyingbacteriabyspecificinhibitingagents 2-Chloro-6 ( trichloromethyl) pyridine (TCMP) orAllylthiourea (ATU). D.Akgul

  21. BOD Test • Method is based on determination of D.O. DirectmethodDilutionmethod D.Akgul

  22. DirectMethod : IfBOD 5 < 7 mg/L  no dilutionrequired. Ex: Riverwaters Adjustthesampleto20 °C , aeratewithdiffusedairtoincreaseordecreasedissolvedgascontenttonearsaturation. Fill BOD bottles. Measure D.O. İmmediately in firstbottle. Incubate 5 day Measure D.O.  BOD =DO1-DO2 D.Akgul

  23. DilutionMethod Rate at whichoxygen is used in dilution of wastes is in directratiotopercent of waste in dilution. 10% dilutionusesoxygen at 1/10 the rate of a 100% sample. D.Akgul

  24. Controlallenvironmentconditions in thebioassay test. • Free of toxicmaterials • FavourablepHandosmoticcond. • Presence of availablenutrients • Standard Temp. • Presence of mixedorganisms of soilorigin D.Akgul

  25. Industrialwastesmay be free of m.org. Andnutrients • Domesticww Contain org. Nutrients N, P, • Dilutionwaterused in BOD mustcompensatethesedeficiences. Dilutionwater Naturalsurfacewater Tap water=>possibility of toxicityfromchlorineresiduals. Syntheticdilutionwaterpreparedfromdistilled, demineralised w.***thebest Dilutionwatermust be free of toxicsubs. Chlorine, Chloroamines, copper. D.Akgul

  26. Seeding: (maintainnecessarymicroorganism) • Domesticwastewaters Blank (dilutionwater) Dilutedsample Seed Nutrient pH buffer Nitrificationinhibitor Dilutionwater D.Akgul

  27. Dilutions of wastes Set threedilutions Ifstrenght is known twodilutions Somecasesupto 4dilutions • Samplesshoulddeplete at least 2 mg/L D.O. • At least 0.5 mg/L of D.O. remaining at theend of incubation. DO1-DO2=2-7 mg/L D.Akgul

  28. Example 3:Calculatethepercentage of wasteshould be addedto a BOD bottleifthe BOD range is; 200/2=100 100/100=1 % 40/2=20 100/20=5 % 20,000/2=10,000 100/10000=0.01% D.Akgul

  29. Incubationbottles: • Glassstoppers • Preventairtrappingduringincubation • Waterseal. • Bottlesshould be free of organicmatter. • Cleanwithchromicacidordetergent. • Rinsecarefully • Allcleaningagentsremovedfrombottle • 4 rinsewith tap water, • final rinsewithDistilledor deionized water D.Akgul

  30. Calculation of BOD BOD(mg/L)= DOb=DO found in blank at theend of incubation DOi=DO found in dilutedsample at theend of incubation DOs= DO thatoriginallypresent in undilutedsample D.Akgul

  31. Example 4: Calculate BOD concentration in mg/L D.Akgul

  32. Respirometricmeasurement OXYTOP Respirometricmethod:Respirometric methods provide direct measurement of the oxygen consumed by microorganisms from an air or oxygen-enriched environment in a closed vessel under conditions of constant temperature and agitation. D.Akgul

  33. Rate of BiochemicalOxidations Differs in diff. Wastes D.Akgul

  34. D.Akgul

  35. D.Akgul

  36. Effluentfrombioligicalwwtpdifferentfromrawwastes • Nature of the org. Matter • Theability of theorganismspresenttoutilizethe org. Matter • Rate of hydrolysisanddiffusion • Ex: Glucose high k • Lignin, Syntheticdetergents  slowlyattackedbybacteria D.Akgul

  37. BODLandTheoreticalOxygenDemand • Total BOD or L = ThODconsideredto be equal • Oxidation of glucose C6H12O6 + 6O2 6CO2+ 6H2O 180 192 D.Akgul

  38. 192 g O2 /mole of glucose OR 1.065 mg O2 /mg glucose 300 mg/L of glucose ThOD= Experimentally; BOD measurements (20 day) BOD(L)= 250-285 mg/L 85% of theoreticalamount Not alltheglucoseconvertedto CO2andwater 300*192/180=320 mg/L D.Akgul

  39. Org. Matter Foodmaterial  • Energy • Growth • Reproduction Part of org. Matter  Convertedtocelltissue Celltissuewillremainunoxidizedtillendogeneousrespiration • Whenbacteriadietheybecomefoodmaterialforothers. • FurthertransformationtoCO2 , H2O andcelltissue • Livingbacteria + Deadones  Foodforhigherorganisms  Protozoans D.Akgul

  40. A certainamount of organicmatterremains in thesetransformations. • Resistanttofurtherbiologicalattack. • Humus amount of org. mattercorrespondingtodiscrepancybetween total BOD andThOD. D.Akgul

  41. Analysis of BOD Data Calculation of k value is neededtoobtain L using BOD5. k and L aredeterminedfrom a series of BOD measurements. Methods: • Leastsquares • Thomas Method • Methods of moments • Daily-differencemethod • Rapidratiomethod • FujimotoMethod D.Akgul

  42. Leastsquaresmethod: Fitting a curvethroug a set of data points. Sum of thesquares of theresidualmust be minimum. (Differencebetweenobservedandthefittedvalue) D.Akgul

  43. dy/dt (t=n) = k (L-yn) • Forthe time series of BOD measurements on thesamesampleEqn. May be writtenforeach of thevarious n data points. • k and L  unknown D.Akgul

  44. If it is assumeddy/dtrepresenttheslope of thecurveto be fittedthroughall data pointsfor a given k and L value, twosideswill not be equalbecause of experimentalerror. Difference R R=k(L-y)-dy/dt R=kL-ky-y’ a=kL -b=k R=a+by-y’ D.Akgul

More Related