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:

;

#### Therefore,

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.,       =

#### 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, , 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