WASTEWATER
TREATMENT-VI
SOLUTION
1. Explain the critical role of hydrogen oxidizing
methanogens (HOM) in anaerobic treatment. (2)
Answer:
During
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)
Answer:
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)
Answer:
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)
Answer:
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