Download CE: 461

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) = 6.67 x 10 0.015.(3520)  3.97 /d
Ks(20oC) = 2224 x 100.046.(3520)  10892 mg/L
k
0.015
Putting  = 0, q  d 
 0.375 /d
YT
0.04
3.97.S
q  0.375 
,
or, (0.375).(10892) = (3.97- 0.375).S
10892  S
or,
S = 1136 mg/L
qm(45oC) = 6.67 x 10 0.015.(3545)  9.42 /d
Ks(45oC) = 2224 x 100.046.(3545)  771.14 mg/L
k
0.015
Putting  = 0, q  d 
 0.375 /d
YT
0.04
9.42.S
q  0.375 
,
or, (0.375).(771) = (9.42- 0.375).S
771  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 (  c ) 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 NO3 concentration (as N), and the total
oxygen requirement for both carbon oxidation and nitrification.
Data:
KS = 40 mg/L;
(KS)N = 2.0 mg/L;
qm = 4 /d;
(qm)N = 2 /d
YT = 0.5;
Kd = 0.05 /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:
1

 0.1 /d ;
 c  10 days ;
c
q
q m .S
4.S
Therefore,

 0.3 ;
K S  S 40  S
Also, V  .Q  8 .10000  3333 m 3
24
q
Hence, X 
  K d (0.1  0.05)

 0.3 /d
YT
0.5
S = 3.24 mg/L
(So  S) (600  3.24)

 5968 mg/L
8 .0.3
.q
24
Also, X  .X.V 
(0.10).(5968).(3333).(1000)
 1989 Kg/d
10 6
Oxygen Requirement (in Kg/d)
1.5.Q.(So – S) –1.42.(X)
(600  3.24)
= 1.5. 10 7.
 1.42.(1989)  6127 Kg/d
10 6
=
Formula weight of biomass, C5H7O2N = 5(12)+7(1)+2(16)+1(14) = 113
14
.1990  246.42 Kg/d Nitrogen
Therefore, production of 1990 Kg/d biomass requires:
113
40.10 7
Influent Nitrogen concentration = 40 mg/L (as N), or
 400 Kg/d
10 6
Therefore, Nitrogen available for nitrification = 400 – 246.42 = 153.57 Kg/d
Or,
[TKN]i = 15.36 mg/L (as N)
Nitrification Calculations:
() N  0.1/d ;
Also, (q) N 
(q) N 
() N  (K d ) N 0.1  0.05

 0.75 /d
(YT ) N
0.2
(q m ) N .[TKN ]
;
(K S ) N  [TKN ]
Therefore, [TKN ] 
Effluent TKN =
Effluent Nitrate =
[(q) N .(K S ) N ]
(2).0.75

 1.2 mg/L
[(q m ) N  (q) N ] 2  0.75
(15.36 - 1.2) =
1.2 mg/L (as N)
14.16 mg/L (as N)
[TKN ]i  [TKN ] 15.36  1.2
42.72
 57 mg/L

 42.72 ;i.e., X N 
0.75

0.33
(3333.103 )
X N   N .X N .V  0.1.(57).
 19 Kg/d
10 6
Also, (q) N .X N 
Additional Oxygen Requirement (Kg/d) = 4.57.Q. { [TKN ]o  [TKN ] }  1.42.(X N )
(15.36  1.2)
i.e.,
= 4.57.107.
 1.42.(19)  617 Kg/d
106
Total Oxygen Requirement = 6127 + 617 = 6744 Kg/d
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, SO 4   8e  8H   S   4.H 2O , 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.,
10.106.800
 8000 Kg/d
106
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
16
Methane Production = 5600.  1400 Kg/d
64
16 Kg of methane at STP = 22.4 m3
1400 Kg of methane at STP =
1400
.22.4  1960 m3
16
Therefore Methane Production = 1960 m3/d