Environmental Engineering Lecture 4b

The environmental engineer must have considerable knowledge of the biological characteristics of waste water because it is a very important factor in wastewater treatment. The Engineer should know: 1.The principal groups of microorganisms found in wastewater. 2.The pathogenic organisms. 3.Indicator for the presence of pathogens 4.The methods to evaluate the toxicity of treated waste water Why Biology

Bacteria  Description One celled plants Colorless Self reproduction (binary every minutes) Use soluble food Different shapes 0.5 – 5 micrometer  Role Decaying organic matter in nature Stabilizing organic waste in plants

Bacteria  Classifications (based on source of nutrients) A.Heterotrophic (Use organic matter for energy and synthesis 1.Aerobic (requires free DO to decompose OM) 2.Anaerobic (oxides OM by the using Oxygen bound in other compounds such as Nitrates and sulfates)

Bacteria 3.Facultative (can live in both environments ) Example (Nitrification – De-nitrification)

Bacteria Energy use:

Bacteria  Classifications (based on source of nutrients) B.Autotrophic (Use inorganic matter for energy and synthesis 1.Sulfur bacteria Leads to corrosion in sewers 1.Iron bacteria (converts soluble ferrous to insoluble ferric)

Some Important Bacteria PseudomonasReduce NO 3 - to N 2, So it is very important in biological nitrate removal in treatment works. ZoogloeaHelps through its slime production in the formation of flocs in the aeration tanks. Sphaerotilus natunsCauses sludge bulking in the aeration tanks. Bdellovibriodestroy pathogens in biological treatment. AcinetobacterStore large amounts of phosphate under aerobic conditions and release it under an – anaerobic condition so, they are useful in phosphate removal. Nitrosomonas transform NH 4 + into NO 2- Nitrobactertransform NO 2- to NO 3- Coliform bacteriaThe most common type is E-Coli or Echerichia Coli, (indicator for the presence of pathogens). E-Coli is measured in (No/100mL)

Protozoa  Description Single celled animals Binary reproduction Complex digestive system Use solid organic matter as food Aerobic (activated sludge – oxidation ponds) Example (Amoeba, Giardia)  Role Feed on bacteria so they help in the purification of treated waste water. Some of them are pathogenic amoeba

Viruses  Description Intracellular parasites Requires living host No metabolic system 20 – 100 nanometer Major health hazards

Algae  Description Photosynthetic plants Simple plant No roots, stem, or leaves Green color Autotrophic

Algae  Role Cause eutrophication phenomena (negative effect) Useful in oxidation ponds. (positive effect) Cause taste problems when decayed. (negative effect)

Infectious agents present in raw domestic wastewater OrganismDiseaseRemarks Bacteria Escherichia coliGastroenteritisDiarrhea Legionella pneumophilaLegionellosis LeptospiraLegionellosisJaundice, fever (weil's disease) Salmonella typhiTyphoid feverHigh fever, diarrhea, ulceration of small intestine. SalmonellaSalmonellosisFood poisoning ShigellaShigellosisBacillary dysentery Vibrio choleraeCholeraExtremely heavy diarrhea, dehydration Yersinia enteroliticaYersinosisDiarrhea

OrganismDiseaseRemarks Viruses Adenovirus (31 types)Respiratory disease Enteroviruses (67 types)Gastroenteritlis, heart Hepatitls AInfectious hepatitisJaundice, fever Norwalk agentGastroenteritlisVomiting ReovirusGastroenteritlis RotavirusGastroenteritlis Infectious agents present in raw domestic wastewater

OrganismDiseaseRemarks Protozoa Balantidium coliBalantidiasisDiarrhea, dysentery CryptosporidiumCryptosporidiosisDiarrhea Entamoeba histolyticaAmebiasis (amoebic dysentery)Prolonged diarrhea with bleeding, abscesses of the liver and small intestine Giardia lambliaGiardiasisMild to severe diarrhea, nausea, indigestion Infectious agents present in raw domestic wastewater

OrganismDiseaseRemarks Helminth Ascaris lumbricoldesAscariasisRoundworm infestation Enterobius vericularisEnterobiasisPinworm Fasciola hepaticaFasciollasisSheep liver fluko Hymenolepis nanaHymenolepiasisDwart tapeworm Teenia saginateTaeniasisBeef tapeworm T. soliumTaeniasisPork tapeworm Trichuris trichiuraTrichuriasisWhipworm Infectious agents present in raw domestic wastewater

Measurements of organic matter  1- Biochemical oxygen demand (BOD): Biochemical oxygen demand (BOD) is the most commonly used parameter to define the strength of a municipal or organic industrial wastewater. The BOD test is used to determine the relative oxygen requirements to treated effluents.

Measurements of organic matter

The following are the most common used methods: 1- Biochemical oxygen demand (BOD): BOD5 is the oxygen equivalent of organic matter. It is determined by measuring the dissolved oxygen used by microorganisms during the biochemical oxidation of organic matter in 5 days 2- Chemical oxygen demand (COD): It is the oxygen equivalent of organic matter. It is determined by measuring the dissolved oxygen used during the chemical oxidation of organic matter in 3 hours.

3- Total organic carbon (TOC) This method measures the organic carbon existing in the wastewater by injecting a sample of the WW in special device in which the carbon is oxidized to carbon dioxide then carbon dioxide is measured and used to quantify the amount of organic matter in the WW. This method is only used for small concentration of organic matter. 4- Theoretical oxygen (ThOD) If the chemical formula of the organic matter existing in the WW is known the ThOD may be computed as the amount of oxygen needed to oxidize the organic carbon to carbon dioxide and other end products. Measurements of organic matter

If L 0  or (BOD ultimate ) or UBOD. L t = L 0 e -kt (BOD remained). BOD t = L 0 - L t = L 0 – L 0 e -kt = L 0 (1-e -kt ) BOD 5 = L 0 (1-e -k5 ) K = 0.23d -1 usually k T = k 20  T-20  = or as given

Determine the 1-day BOD and ultimate BOD for a wastewater whose 5-day 20C BOD is 200 mg/L. The reaction constant K= 0.23d -1 what would have been the 5-day BOD if it had been conducted at 25C? Solution:-  BOD t = UBOD (1-e -kt ) =L 0 (1-e -kt ) 200 = L 0 (1-e -0.23x5 ) L 0 = 293 mg/L (this is UBOD)  Determine the 1-day BOD:- BOD t = L 0 (l-e -kt ) BOD 1 = 293 (l-e -0.23x1 ) = 60.1 mg/L ’  Determine the 5-day BOD at 25C:- K T = K 20 (1.047) T-20  K 25 = 0.23 (1.047) BOD 5 = L 0 (l-e – kt ) = 293 (l-e -0.29x5 ) = 224 mg/L

ThOD: Calculate the Theoretical Oxygen Demand (ThOD) for sugar C 12 H 22 O 11 dissolved in water to a concentration of 100 mg/L. Solution:- C 12 H 22 O O 2  12 CO H 2 O ThOD = ThOD = mg O 2 / L