1. Explain the critical role of hydrogen oxidizing methanogens (HOM) in anaerobic treatment. (2)
anaerobic treatment, obligate hydrogen producing acetogens (OHPA) produce
hydrogen during their respiration. This
hydrogen, when present in high concentrations, is toxic to OHPA. Thus an agent for removal of hydrogen from
the system is required. The hydrogen
oxidizing methanogens (
2. Why treatment of domestic wastewater is impractical in a suspended growth anaerobic (i.e., anaerobic contact) process. (2)
Domestic wastewater treatment requires that the treated wastewater have very low COD, i.e., below discharge standards. Bio-kinetics of anaerobic bacteria demand that to get very low effluent COD, the specific substrate utilization and hence specific growth of these microorganisms be maintained at very low values.
Since, Substrate Utilization Rate, to maintain a reasonably high value of , either q or X has to be increased. Since as stated before, q has to be very low, X has to be increased. Increasing X to very high values is however not very practical, since anaerobic bacteria has very poor settling characteristics, and hence settling in the secondary tank is not very efficient. This results in constant escape of biomass from the system.
Failure to maintain high value of X in the system results in increase in specific substrate utilization rate (q), and resultant high values of Se.
3. Paracoccus denitrificans is a ‘hydrogen oxidising, autotrophic, denitrifying (HOD)’ microorganism. Based on the above characterization, what is the food source, energy source and terminal electron acceptor for this microorganism. (2)
Food Source: Inorganic carbon, i.e., bicarbonate, converted to cell material
Energy Source: Hydrogen gas, oxidized to water
Electron Acceptor: Nitrate ions, reduced to nitrogen gas
4. 10 MLD of wastewater with influent COD (So) of 800 mg/L is treated in an UASB reactor. 80% COD removal was reported. Calculate theoretical methane production, i.e., liters of methane produced per day at STP. Neglect COD conversion to anaerobic bio-mass.
Methane Oxidation Equation: (4)
Q = 10 MLD; So = 800 mg/L; Therefore, Se = 0.2.So = 160 mg/L
Influent COD Loading = Q.So =
Effluent COD Loading = Q.Se =
Therefore, COD transferred to methane = (8000 – 1600) = 6400 Kg/d
COD of methane can be calculated from the following equation:
i.e., 48 Kg of methane will exert an oxygen demand of 192 Kg
Thus, 6400 Kg of oxygen demand is equivalent to methane
48 g methane has a volume of (22.4 x 3) = 67.2 liters at STP
Therefore, 1600 Kg methane has a volume of