EXAMINATION-I
Time:
180 Minutes Full
Marks: 90 (+10 bonus)
Question
1
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. |
S_{o}, mg/L (influent BOD_{5}) |
_{}, days |
S, mg/L (Effluent BOD_{5}) |
X, mg/L (Biomass) |
1. |
250 |
3 |
10 |
120 |
2. |
250 |
1 |
45 |
110 |
Based on the above data
calculate q_{m}, K_{d}, K_{s},
and Y_{T}. (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 BOD_{5}
(S_{o}) = 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 BOD_{5 }(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 BOD_{U} from the pond.
Data:
·
K = 0.1 L/mg/d,
where K is the
first order microbial substrate utilization rate (_{})
·
Y_{T} = 0.5 mg/mg; K_{d} = 0.05 /d (based on BOD_{5})
·
Formula for microbial biomass: C_{60}H_{87}O_{23}N_{12}P
·
Average intensity of solar radiation: 150 calories/cm^{2}/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.(S_{o} - 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 BOD_{u} = 1.42.(28) =
39.76 mg/L
Also,
3550 Kg of algae has BOD_{u} of 4416 Kg
Effluent
algae concentration = 150 mg/L, corresponding BOD_{u} = _{}mg/L
Effluent
soluble BOD_{u} = 1.5.(5) = 7.5 mg/L
Therefore,
total effluent BOD_{u} = 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/m^{3}; Dry Density of Sludge Solids = 2200
Kg/m^{3}
COD reduction efficiency of sludge digester = 60 %; _{};
For sludge digestion, Y_{T} = 0.06 mg/mg,
and K_{d} = 0.03 /d on COD basis
Assumptions:
Solution:
Consider
1 m^{3} 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: r_{l} = 1000 kg/m^{3}; r_{s} = 2200 kg/m^{3}
So, _{}
Therefore, Weight of solids in 1 m^{3} of
sludge = (0.04).(1).(1022.305) =40.89 kg.
Solids Concentration (X_{o})
= 40890 mg/L
Let
the discharge of thickened sludge be Q_{sl}
Therefore, assuming 100 percent solids capture,
Q_{sl}. (40890) = Q_{w}.
(13000) or, Q_{sl} = _{}
Solids
Loading (DX) = _{}
Digester Design:
_{}; Digester Volume
(V_{d}) = _{} _{}; _{}
_{}
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 CaCO_{3}. Discharge
standards for wastewater are, COD: <50 mg/L; TKN: < 2 mg/L; Total-P: <
1 mg/L; pH: 6.5–8.5. 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-à-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
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 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.
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 m^{3}/m^{2}/hr before the terminal
head-loss of 3 m was reached. Filter
backwashing rate was 1 m^{3}/m^{2}/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 m^{3}/m^{2}/h
Filter
is off-line for 1.5 hours every 24 hours
Therefore,
effective filtration rate = _{} m^{3}/m^{2}/h
Therefore,
required filter cross-sectional area = _{} m^{2}
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 m^{2}
Corrected
actual filtration rate = _{}m^{3}/m^{2}/h
Filtered
water required for backwashing = 1.(5).(45.1) .3 = 676.5 m^{3}/d
Hence
total filtered water production = 10 – 0.676 = 9.324 MLD
Consider water from a polluted river having BOD_{5}
= 5 mg/L, TKN = 1 mg/l (as N), and MPN: 10^{6} 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 Cl_{2})
required per liter of this water (consider both pre and post-chlorination) such
that after treatment BOD_{5}, TKN, NH_{3}-N are negligible and
MPN < 1organism/mL.
Assumptions:
Solution:
Chlorine dose required during pre-chlorination for
destruction of BOD_{5} = 5 mg/L as Cl_{2}
All TKN in water is converted to NH_{3}-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 = 10^{6}
/mL
Removal during water treatment up to
post-chlorination = 2 Log
Hence microorganism concentration just before post
chlorination = 10^{4} /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 Cl_{2}
Therefore, total chlorine dose required = 5 + 7.6 +
1.6 = 14.2 mg/L as Cl_{2}