EXAMINATION-III
SOLUTION
The total number of marks allotted for various questions add up to
35. However, the marks obtained will be
reported on 30.
1. A. Describe the
difference between the recycling in the activated sludge and trickling filter
processes. (2)
Solution:
The settled sludge is recycled in ASP, while the
treated effluent is recycled in TF. The
objective of recycling in ASP is to maintain the desired biomass concentration
in the aeration tank, while in TF, recycling allows the HLR to be varied
independent of the OLR, thus allowing uniform wetting of the filter media.
B. Explain
why anaerobic reactors may fail if organic loading rate (OLR) is increased
suddenly. What are the early warnings of
reactor failure and how can such failure be prevented. (2)
Solution:
If OLR in an anaerobic reactor is increased suddenly,
the acid formers will become more active, and VFA concentration in the reactor
will increase. If the buffering capacity
in the reactor is insufficient, all alkalinity in the reactor will be exhausted
due to the increased acid formation, and reactor pH will go down. At reactor pH<7, the methane formers will
become inactive, and methane production will thus stop. This will stop COD reduction in the
reactor. Reactor failure in such
circumstances may be avoided by adding a buffering agent like soda (Na2CO3)
to the reactor to stop decline in pH.
C. Calculate
the minimum effluent COD that can be expected from an anaerobic reactor at 45oC
and 20oC, assuming the microbial kinetics of anaerobic reactors may
be described using the following relationships, YT = 0.040; Kd
= 0.015 /d; q = 6.67 x 10-0.015.(35-T) /d;
Ks = 2224 x 100.046.(35-T) mg/L. ‘T’ is the temperature expressed in degree
Celsius. (2)
Solution:
qm (20oC) = 3.97 /d
Ks(20oC) = 10892 mg/L
Putting m = 0,
, or, (0.375).(10892) = (3.97- 0.375).S
or, S =
1136 mg/L
qm(45oC) = 9.42 /d
Ks(45oC) = 771.14 mg/L
Putting m = 0,
, or,
(0.375).(771) = (9.42- 0.375).S
or, S = 32
mg/L
D. Explain
why a suspended growth anaerobic reactor gives very poor COD removal when used
for treatment of domestic wastewater. (2)
Solution:
Anaerobic reactors must be maintained at very low
specific growth rates for obtaining low effluent COD values. This means that very high biomass
concentration must be maintained in the reactor to achieve reasonable rates of
substrate utilization. In suspended
growth systems, maintenance of high biomass concentration is only possible if
the biomass escaping from the reactor is settled and recycled back
efficiently. However this is difficult
in anaerobic reactors because anaerobic sludge has poor settling characteristics. Also, biomass escaping with the treated
effluent degrades the effluent quality.
E. In a single stage aeration tank designed for carbon
oxidation and nitrification, the dissolved oxygen concentration has to be
maintained at 2.5-3.0 mg/L, as opposed to value of 1.0 mg/L in cases where only
carbon oxidation is required. Why? (2)
Solution:
In single stage nitrification systems typically only
1-2 percent of the biomass consists of nitrifying organisms. Often such organisms are enmeshed in clusters
of heterotrophic organisms responsible for carbon oxidation, and thus have to
compete for aqueous oxygen. Under the
circumstances, high DO concentration must be maintained in such reactors to
ensure that aqueous oxygen diffuses into the inner parts for biomass clusters
where many nitrifying microorganisms may be found.
2. Consider a wastewater flow of 10 MLD with a BOD5
value of 600 mg/L. It is proposed to
treat this waste using an activated sludge process (ASP). The size of the proposed aeration tank will
such that the water is detained for 8 hours.
Complete mixing is assumed in the aeration tank. The BSRT () of the system is 10 days.
In addition to the BOD5, this water also contains nitrogen
and phosphorus. The nitrogen
concentration expressed as Total Kjeldahl Nitrogen (TKN) is 40 mg/L (as N), and
phosphorus concentration is not limiting.
Assuming that nitrification occurs in this system, calculate the
effluent TKN and concentration (as N),
and the total oxygen requirement for both carbon oxidation and nitrification.
Data:
KS
= 40 mg/L; qm
= 4 /d; YT = 0.5; Kd = 0.05 /d;
(KS)N =
2.0 mg/L; (qm)N
= 2 /d (YT)N =
0.2 ; (Kd)N =
0.05/d
Assumptions:
·
Biomass may be
represented as C5H7O2N
·
Nitrogen incorporation in heterotrophic biomass must be accounted for.
·
Nitrogen incorporation in autoprophic biomass may be neglected.
(15)
Solution:
Carbon Oxidation Calculations:
; ;
; Therefore, S = 3.24 mg/L
Also,
Hence,
Also,
Oxygen Requirement (in
Kg/d) = 1.5.Q.(So – S) –1.42.(DX)
= 1.5.
Formula weight of biomass,
C5H7O2N = 5(12)+7(1)+2(16)+1(14)
= 113
Therefore, production of
1990 Kg/d biomass requires:
Influent Nitrogen
concentration = 40 mg/L (as N), or
Therefore, Nitrogen
available for nitrification = 400 – 246.42 = 153.57 Kg/d
Or, [TKN]i = 15.36 mg/L (as N)
Nitrification Calculations:
;
Effluent
TKN = 1.2
mg/L (as N)
Effluent
Nitrate = (15.36 - 1.2) = 14.16 mg/L (as N)
Also, ;i.e.,
Additional
Oxygen Requirement (Kg/d) = 4.57.Q.
{}
i.e.,
=
3. 10 MLD of wastewater with
influent COD (So) of 800 mg/L is treated in an UASB reactor. In addition, the influent wastewater contains
300 mg/L of sulfate, and negligible amounts of sulfide. 50% conversion of this sulfate to sulfide, as
per the following equation, , was reported due to the action of sulfur reducing
bacteria**. In addition, 70% COD removal
was reported. Calculate theoretical
methane production, i.e., volume of methane produced per day at STP. Neglect
COD conversion to anaerobic biomass.
** These are anaerobic bacteria
using organic acids produced by acid-producing bacteria under anaerobic
conditions as food and energy source and sulfate as the terminal electron
acceptor. Sulfate reducing bacteria are present in anaerobic reactors if
sulfate concentration is high. (10)
Solution:
Influent COD
= 800 mg/L, i.e.,
Influent COD
(Liquid Phase) =
Methane COD (Gas Phase) +
Sulfide COD (Liquid Phase) +
Effluent COD (Liquid
Phase) + Biomass COD (Solid Phase)
Biomass COD
(Solid Phase) is neglected.
Sulfide COD
(Liquid Phase) + Effluent COD (Liquid Phase) = 0.30.(8000) = 2400 Kg/d
Therefore Methane
COD (Gas Phase) = 8000 – 2400 = 5600 Kg/d
64 Kg of
Methane COD = 16 Kg of Methane
Methane
Production = Kg/d
16 Kg of
methane at STP = 22.4 m3
1400 Kg of
methane at STP =
Therefore
Methane Production = 1960 m3/d