EXAMINATION-I

 

 

Time: 180 Minutes Full Marks: 90 (+10 bonus)

 

Question 1

Sketch the top and front view of a rectangular primary sedimentation tank, clearly showing the inlet and outlet arrangements and sludge and scum collection mechanisms. Label various relevant features. Also draw expanded and/or sectional views of the inlet and outlet regions of the tank to show all details and label all relevant features.

Solution: No solution provided.

 

Question 2

a.       To predict the performance of an activated sludge unit, one must have data on microorganism properties. To obtain this data, a laboratory experiment is conducted using a continuous flow completely mixed reactor without recycle. Data obtained after two runs are as follows:

Run No.

So, mg/L

(influent BOD5)

, days

S, mg/L

(Effluent BOD5)

X, mg/L

(Biomass)

1.

250

3

10

120

2.

250

1

45

110

Based on the above data calculate qm, Kd, Ks, and YT. (8)

b. Also determine the minimum value of for this reactor, below which no substrate removal shall occur. Explain the physical significance of this minimum value. (2)

 

Solution:

We know,

Hence, , and

 

We also know, ,

Hence, , or,

also, or,

 

Solving, ;

 

Also, , Hence,

Also,

 

Hence,

 

Solving,

 

When there is no substrate removal in the reactor,

Hence,

Also,

Hence,

 

Physical Significance:

If for this reactor is less than , then no substrate removal will occur because biomass will flow out of the reactor at a faster rate than it can be produced. Under such conditions, the biomass concentration in the reactor at steady state will be zero.

 

Question 3

Draw a neat sketch of a trickling filter (plan and sectional elevation) and label all the essential parts. Write an essay on trickling filter operation, covering the following points, mode of biomass growth and removal from the reactor, mode of substrate utilization and oxygen uptake and importance of OLR, HLR, and recycling. (3 + 7)

 

Solution: No solution provided.

 

Question 4

Draw a neat sketch (sectional elevation) of an Up-flow Anaerobic Sludge Blanket (UASB) reactor and explain its various parts and their operation. Explain why and how this reactor configuration is the more successful in treating domestic wastewater as compared to other anaerobic reactor configurations. (5 + 5)

 

Solution: No solution provided.

 

Question 5

1 MLD of wastewater with influent soluble BOD5 (So) = 75 mg/L, TKN = 13 mg/L (as N) and 2 mg/L phosphorus (as P) is to be treated in an oxidation pond such that effluent soluble BOD5 (S) is 5 mg/L. Calculate the oxidation pond surface area, assuming the depth of the pond to be 0.5 m. Calculate oxygen requirement for microbial respiration and oxygen supplied by algal growth, and check adequacy of the design by assuming that 50 percent of the oxygen produced by algal growth is available for microbial respiration. Neglect oxygen input into the pond by mass transfer from atmosphere. Calculate the effluent algal, effluent nitrogen and phosphorus concentration from the pond, and also calculate the total effluent BODU from the pond.

Data:

                     K = 0.1 L/mg/d,

where K is the first order microbial substrate utilization rate ()

                     YT = 0.5 mg/mg; Kd = 0.05 /d (based on BOD5)

                     Formula for microbial biomass: C60H87O23N12P

                     Average intensity of solar radiation: 150 calories/cm2/d

                     Solar energy utilization efficiency for algae: 6 percent

                     Energy content of algal bio-mass: 6000 calories/g algae

                     Equation for algal photosynthesis:

 

Solution:

We know, ;

Therefore,

Also,

Also, , or,

Sludge Production () = Q.X =

 

Oxygen Requirement = 1.5Q.(So - S) 1.42(DX) = 1.5.(70) (1.42).(28) = 65.24 Kg/d

 

Volume of Oxidation Pond =

Assuming depth to be 0.5 m, Surface Area (A) =

Algae production =

Total algal production = (15).(10000) = 150 kg/d

 

Assuming 1.3 Kg oxygen production per Kg algal production,

Oxygen Production = (1.3).(150) = 195 Kg/d

 

Oxygen available for microbial respiration = (0.5).(195) = 97.5 Kg/d

 

Since oxygen available is more that oxygen requirement for microbial respiration, the design is adequate.

 

Biomass production = 28 Kg/d

Formula weight of biomass = 60.(12) + 87.(1) + 23.(16) + 12.(14) + 1.(31) = 1374

Hence, 1374 Kg of biomass contains 168 Kg of nitrogen and 31 Kg of phosphorus

28 Kg of biomass contains kg nitrogen and kg of phosphorus

Algae production = 150 Kg/d

Formula weight of algae = 106.(12) + 263.(1) + 110.(16) + 16.(14) + 31 = 3550

Hence, 3550 Kg of algae contains 224 Kg of nitrogen and 31 Kg of phosphorus

150 Kg of algae contains kg nitrogen and kg of phosphorus

Therefore, total nitrogen uptake = 3.42 + 9.46 = 12.88 Kg/d

Total phosphorus uptake = 0.63 + 1.31 = 1.95 Kg/d

Influent nitrogen loading = Kg/d

Influent phosphorus loading = Kg/d

Hence effluent dissolved nitrogen concentration = 13-12.88 = 0.12 mg/L (as N)

Effluent dissolved phosphorus concentration = 2 1.95 = 0.05 mg/L (as P)

 

Effluent biomass concentration = 28 mg/L, corresponding BODu = 1.42.(28) = 39.76 mg/L

Also, 3550 Kg of algae has BODu of 4416 Kg

Effluent algae concentration = 150 mg/L, corresponding BODu = mg/L

Effluent soluble BODu = 1.5.(5) = 7.5 mg/L

Therefore, total effluent BODu = 39.76 + 186.59 + 7.5 = 233.85 mg/L

 

Question 6

0.25 MLD of sludge is generated from the activated sludge process of a wastewater treatment plant. The solids concentration in this sludge is 13,000 mg/L. This sludge is first processed in a sludge thickener, where the solids content is increased to 4 percent (weight basis). Next the sludge is treated in the anaerobic digester for reduction in sludge solids. The digested sludge is then applied to sludge-drying beds for reduction of water content to 55%. Determine the weight and density of dried sludge that will be produced per day.

Density of Water = 1000 Kg/m3; Dry Density of Sludge Solids = 2200 Kg/m3

COD reduction efficiency of sludge digester = 60 %; ;

For sludge digestion, YT = 0.06 mg/mg, and Kd = 0.03 /d on COD basis

Assumptions:

 

Solution:

Consider 1 m3 of sludge after thickening

Let the sludge density be

Weight of this sludge =

Weight of solids in the sludge = ; Volume =

Weight of liquid in the sludge = ; Volume =

Therefore, Total Volume = [+ ] = 1

Given: rl = 1000 kg/m3; rs = 2200 kg/m3

 

So,

 

Therefore, Weight of solids in 1 m3 of sludge = (0.04).(1).(1022.305) =40.89 kg.

Solids Concentration (Xo) = 40890 mg/L

 

Let the discharge of thickened sludge be Qsl

Therefore, assuming 100 percent solids capture,

Qsl. (40890) = Qw. (13000) or, Qsl =

Solids Loading (DX) =

 

Digester Design:

; Digester Volume (Vd) = ;

Assuming 60% treatment efficiency,

Also, , Also,

Therefore,

Anaerobic Sludge Production (=

 

Therefore, Reduction in Solids =

Therefore, Solids Loading to Sludge Drying Beds = (3250-1891) = 1359 kg/d

 

 

After Drying:

Therefore, Dried Sludge Production Rate =

 

Density of Dried Sludge =

 

Question 7

A food-processing unit produces 0.1 MLD of wastewater with the following characteristics. COD: 10,000 mg/L; TKN: 500 mg/L; Total-P: 100 mg/L; pH: 7.5; Alkalinity: 500 mg/L as CaCO3. Discharge standards for wastewater are, COD: <50 mg/L; TKN: < 2 mg/L; Total-P: < 1 mg/L; pH: 6.58.5. Based on the above information, provide a preliminary sketch of appropriate treatment process train, including those for residuals management. Explain the function of each unit in the process train, and also mention the probable percentage of pollutant reduction in each unit. Use of a mixture of engineered and natural methods for treatment is encouraged.

Solution: This is one of many possible treatment trains.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 


Question 8

Consider settled raw water of three types;

Based on you understanding of water treatment unit operations suggest the best treatment option (in terms of cost) for all three cases to reduce the turbidity to < 2.5 NTU. Explain clearly why the selected option in each case is the best vis--vis other available options.

 

Solution:

Case A: Recommended treatment train is as follows.

Coagulant/Poly-electrolyte addition Rapid Mixing Rapid Sand Filtration

 

Case B: Recommended treatment train is as follows

Coagulant/Ploy-electrolyte addition Rapid Mixing Flocculation

 

Rapid Sand (Dual Media) Filtration

 

Case C: Recommended treatment train is as follows.

Coagulant and Soda addition Rapid Mixing Flocculation

 

Secondary Sedimentation

 

 


Rapid Sand Filtration

In all above cases, excellent effluent quality might be obtained by the following treatment train:

Coagulant (High Dose) and Soda Addition Rapid Mixing

 

Flocculation to Encourage sweep floc formation

 

 


Secondary Sedimentation

 

 


Rapid Sand Filtration

 

However, such a treatment chain is not recommended because of the large chemical cost (coagulant and soda), and the cost of disposing the large quantity of sludge produced.

Case A. Flocculation will be useless in this case due to the low number of particles present, unless sweep floc formation is desired (which is not the case). Consequently, secondary sedimentation will also be redundant. Under the circumstances, direct filtration is the most logical option.

 

Case B. Flocculation will be moderately effective in this case, but flocs large enough to settle efficiently may not be formed, unless sweep-floc formation is desired (which is not the case). Consequently, secondary sedimentation will be redundant. Flocculation will however reduce particle loading on the filter, thus extending filter runs. Dual media filter is proposed since the influent particle size is higher in this case as compared to Case A.

 

Case C. Flocculation will be very effective in this case due to large influent particle concentration. Consequently, secondary sedimentation will also be effective. However on coagulant addition, pH is likely to decline in this case to due the low alkalinity of water. Hence the conventional treatment chain, with soda addition along with coagulant to maintain pH drop, is thought to be the most logical option in this case.

Question 9

10 MLD of water after secondary sedimentation (average turbidity: 10 NTU) is to be filtered through a battery of rapid sand filters to reduce water turbidity to < 2.5 NTU. Based on pilot plant studies, it was determined that 60 cm filter beds of sand (0.5 mm average sand diameter) were suitable for this purpose. It was further determined that such beds could be operated for 7.5 hours at a filtration rate of 10 m3/m2/hr before the terminal head-loss of 3 m was reached. Filter backwashing rate was 1 m3/m2/min and the backwash time was 5 minutes. A filter unit will be off-line for 30 minutes during each backwash operation. Based on this information, determine the numbers of filter units to be provided and dimensions of each unit. Determine how much filtered water is required for backwashing each day and hence determine the filtered water production per day.

 

Solution:

Nominal filtration rate: 10 m3/m2/h

Filter is off-line for 1.5 hours every 24 hours

Therefore, effective filtration rate = m3/m2/h

Therefore, required filter cross-sectional area = m2

Let two filters be provided for this purpose

Let the length of each filter be 5.5 m and width 4.1 m (length: width = 1.34)

Therefore total filter cross-section area provided = 2.(5.5).(4.1) = 45.1 m2

Corrected actual filtration rate = m3/m2/h

Filtered water required for backwashing = 1.(5).(45.1) .3 = 676.5 m3/d

Hence total filtered water production = 10 0.676 = 9.324 MLD

 

Question 10

Consider water from a polluted river having BOD5 = 5 mg/L, TKN = 1 mg/l (as N), and MPN: 106 organisms / mL. This water will be treated in a conventional water treatment plant and supplied for potable purposes. Compute the chlorine dose (in mg/L as Cl2) required per liter of this water (consider both pre and post-chlorination) such that after treatment BOD5, TKN, NH3-N are negligible and MPN < 1organism/mL.

Assumptions:

 

Solution:

Chlorine dose required during pre-chlorination for destruction of BOD5 = 5 mg/L as Cl2

All TKN in water is converted to NH3-N during this process.

Hence ammonia concentration in water before post-chlorination = 1 mg/L (as N)

 

Breakpoint chlorination has to be performed to destroy ammonia in water.

Relevant equation:

Ammonia concentration in water = mmoles/L

Chlorine required for destruction of ammonia = 1.5.(0.0714 ) = 0.1071 mmoles/L,

Therefore, breakpoint chlorination dose = 71.(0.1071) = 7.6 mg/L

 

Initial microorganism concentration = 106 /mL

Removal during water treatment up to post-chlorination = 2 Log

Hence microorganism concentration just before post chlorination = 104 /mL

To get this concentration below 1 /mL, 5 log kills are required

 

C.t for 5 log kills = 96

Contact time = 1 hour = 60 minutes

Therefore required free chlorine residual dose = mg/L as Cl2

Therefore, total chlorine dose required = 5 + 7.6 + 1.6 = 14.2 mg/L as Cl2