The primary clarifier was designed as a long rectangular basin with two channels each having dimensions of 24.5m x 8.16m x 3.3m deep to treat an average flow of 16,000 cubic meters per day. The completely mixed activated sludge plant was sized to treat an influent BOD of 1,200mg/L to an effluent BOD of 200mg/L with a solids retention time of 5 days and MLSS of 5,000mg/L based on given kinetic parameters.

1. Qualifying Examination Presentation
Somdipta Bagchi
A16CE09005
by
Under the supervision of
Dr. Manaswini Behera
Assistant Professor, School of Infrastructure
School of Infrastructure
Indian Institute of Technology, Bhubaneswar

2. Q1. (a). A private solid waste collector wishes a locate a MRF near a commercial area.
The container would like to use hauled container system but fears that the haul cost
might be prohibitive. What is the maximum distance away from the commercial area
that the MRF can be located so that the weekly costs of the hauled container system do
not exceed those of stationary container system? Assume that one collector-driven will
be used with each system and the following data are applicable. For the purpose of this
example assume the travel times t1 and t2 are inclusive in the off route factor.
a. Hauled container system
Number of trips = 56trips/week, container pick up time = 0.033h/trip, container
unloading time = 0.033h/trip, a = 0.022h/trip, b = 0.022h/km, at site time =
0.053h/trip, time to travel between container locations = 0.067 h/trip, over head
cost = Rs. 2400 per week, operation cost = Rs. 900/h of operation, assume 8h
working day and off-route factor = 0.15
b. Stationary customer system
Quantity of solid waste = 300m3/week, container size = 8m3/container, container
utilisation factor = 0.67, collection vehicle capacity = 30m3/trip, compaction ratio =
2, unloading time = 0.05h/trip, a = 0.022h/trip, b = 0.022h/km, at site time =
0.1h/trip, time to travel between container = 0.067h/trip, overhead cost = Rs 4500
per week, operation cost = Rs 1200/h of operation and off route factor = 0.15.

3. Solution :-
a. Hauled container system
The time required per week, Tw , as a function of the round-trip haul distance, can
be determined using the following expression,
Tw(hcs) = Nw(hcs)( Phcs + s + a + b*x )/[ H(1 – W) ]
where,
Nw(hcs) = number of collection trips required per week, trips/week.
Phcs = pickup time per trip for hauled container system, h/trip,
s = at site time per trip, h/trip , H = number of working hours per day, h,
a = empirical constant, h/trip , W = off site factor,
b = empirical constant, h/km ,
x = round trip haul distance, km/trip,
Phcs can be calculated by the following expression,
Phcs = pc + uc + dbc
where,
pc = time required to pick up loaded container, h/trip
uc = time required to unload empty container, h/trip
dbc = time required to drive between container locations, h/trip
thus,
Phcs = pc + uc + dbc
= 0.033h/trip + 0.033h/trip + 0.067h/trip
= 0.133h/trip.

4. therefore,
TW(hcs) = 56trips/wk( 0.133h/trip + 0.053h/trip + 0.022h/trip + 0.022h/km*x)
8h/d(1-0.15)
= (1.713 + (0.181/km)*x)d/wk
Thus, the weekly operational cost as a function of the round trip haul distance can be
determined as
= Rs 900/h *8(h/d) (1.713 + (0.181/km)*x)d/wk
= [ 12333.6 + 1303.2/km * x ] Rs/wk
b. Stationary container system
The time required per week, Tw , as a function of the round-trip haul distance, can be
determined using the same expression,
Tw(scs) = Nw(scs)( Pscs + s + a + b*x )/[ H(1 – W) ]
Here, the pickup time per container can be calculated by
Pscs = Ct(uc) + (np- 1)dbc
where,
Ct = number of containers emptied per trip, containers/trip
uc = time required to unload empty container, h/trip
np = number of container pickup locations per trip, locations/trip
dbc = average time spent driving between container locations, h/location

5. Number of containers emptied per trip can be determined by,
Ct = vr/cf
v = volume of the collection vehicle, m3/trip
r = compaction ratio
c = container volume, m3/container
f = weighted container utilization factor
thus,
Ct = vr/cf
= (30m3/trip * 2)/(8m3/container * 0.67)
= 11.19 container/trip
= 11 containers/trip
and,
Pscs = Ct(uc) + (np- 1)dbc
=(11 containers/trip * 0.05h/trip) + ((11-1) locations/trip * 0.067h/trip)
= 1.22 h/trip
also,
Nw = Vw/vr
where,
Nw = number of collection trips required per week, trips/wk
Vw = weekly quantity of waste collected, m3/wk

6. Nw = Vw/vr
= (300m3/wk)/(30m3/trip * 2)
= 5 trips/wk
Therefore,
Tw(scs) = Nw(scs)( Pscs + s + a + b*x )/[ H(1 – W) ]
= 5 trips/wk(1.22h/trip + 0.1h/trip + 0.22h/trip + 0.22h/km * x)
8(h/d)(1-0.15)
= (0.99 + 0.016/km * x)d/wk
Thus, the weekly operational cost as a function of the round trip haul distance can be
determined as
= Rs 1200/h *8(h/d) (0.99 + (0.016/km)*x)d/wk
= [ 9504 + 153.6/km * x ] Rs/wk
The maximum round trip haul distance at which the cost for hauled container systems
equals the cost for the stationary container systems by equating the total costs for the two
systems by equating the total costs for the two systems and solving for x.
2400(Rs/wk)+[12333.6+1303.2/km*x](Rs/wk) = 4500(Rs/wk)+[9504+153.6/km*x](Rs/wk)
x = 0.6346 km.

7. (b). Estimate the amount of methane that can be recovered from 2000kg of waste,
consisting of organic fraction of MSW. Assume moisture content of the waste is 20%;
volatile solids 93% of total solids; biodegradable volatile solids is 70% of volatile solids,
expected biodegradable volatile solids (BVS) conversion efficiency is 85% and gas
production 12m3/kg BVS destroyed, methane is 65% of total gas production.
Solution:-
According to the question,
Total quantity of waste is 2000kg,
Solid content of waste is 80%,
Volatile solid content is 93%,
Biodegradable volatile solid content is 70%, and
Biodegradable volatile conversion efficiency is 85%.
Therefore, the quantity of biodegradable volatile solid converted is
= 2000*0.80*0.93*0.70*0.85
= 885.36 kg
Now, the quantity of gas produced per kg of BVS destroyed is 12m3 .
Therefore, the total volume of gas produced is = 12*885.36 = 10624.32m3.
And the amount of methane generated is thus = 0.65*10624.32
= 6905.808m3 .

8. 2. What are the advantages and disadvantages of public participation in the process of
EIA? Describe the scoping stages involved in the EIA study of laying down new sewerage
network in a city.
Solution:-
Public participation has the following advantages:-
1. Improved understanding of client expectations and user group needs.
2. Improved agency understanding of conservation issues.
3. Improved agency understanding of the role and contribution of the community.
4. Ability to build community support for a project and to improve stakeholder
relationships.
5. Improved public understanding of the agency’s responsibilities.
6. Improved staff and community technical knowledge.
7. Improved quality of decision-making by agencies.
8. Enhanced and informed political process
9. Greater community advocacy for biodiversity protection.
10. Greater access to community skills and knowledge.

9. Disadvantages of public participation:-
1. Public participation can be time consuming.
2. It can be sometimes expensive.
3. To do it effectively, organizations have to build capacity and train staff.
4. If not done properly, public participation processes can result in, for example, loss of
faith in the agency. A negative experience of the process may lead participants to
have negative perceptions of the outcome, and they may be less likely to participate
in future processes.

10. Scoping stages in EIA study for laying down new sewerage network in a city:-
Scoping is the process of identifying the issues to be considered in the impact
assessment and selecting the appropriate alternatives.
1. Appropriate boundaries of the EIA study
a. Project boundaries: The project boundary includes all lands subject to direct
disturbance from the project and associated infrastructure of the sewerage network.
b. Temporal boundaries: Temporal boundaries of a sewerage network are defined by
the life of the project and duration of its construction, operation and abandonment
phases of the project.
c. Administrative boundaries: These boundaries are time and space limitations
imposed because of administrative or economic reasons.

11. 2. Stakeholders
Stakeholders are defined as all those people and institutions that have an interest in
the successful design, implementation and sustainability of the project. These include
positively and negatively affected people by the project.
For laying down of a sewerage network, the stakeholders can be:-
a. government ministers and agencies
b. town residents at household level
c. private sector
d. construction contractors, local builders
e. local technicians, plumbers
f. unemployed people
g. affected persons and families
h. schools and hospitals

12. 3. Key issues and concerns
In this stage the Valued Environmental Components are identified. VECs for the
project are those environmental attributes associated with the proposed project
development, which are identified based on:
a. Concerns expressed by government, the professional community, and directly-
affected stakeholders
b. EIA terms of reference
c. Review of legislation
d. Consideration of available reference material and literature
e. Previous assessment experience including proposed developments in the Project
study areas .
f. Issues and concerns related to resources traditionally used by indigenous people

13. 4. Impact Identification
a. Increase pollution in receiving river because of discharge of not treated or
partially treated wastewater.
b. Clearance of sites from vegetation, as well as the executive of excavation works
using heavy or inappropriate construction practices and soil protection
measures may accelerated erosion, lead to soil instability and landslides in
sloped areas.
c. Surface and groundwater contamination due to sewerage seepage in the case
of wastewater collection system damage.
d. Soil degradation due to stripping and removal of humus layer.
e. Atmospheric pollution by dust possibly contaminated with other air pollutants
resulting from earthworks, load and unload of raw materials.
f. Odour generation from STP and sludge treatment operations.
g. Construction equipment and other operational activities will generate noise
which can affect workers, population and animals living or moving in the vicinity
of working points
h. Destruction or alteration of the habitats of the flora and fauna species

14. 5. Description of the baseline information
a. Surface water in project area and neighbourhood area-distances to project
location.
b. Nature and location of the aquifers in the project area, water movement direction
on groundwater.
c. Drainage in project area, includes the location and capacity, canals drains and
rivers.
d. Soils and geomorphology
e. Sensitive locations to air quality in project area and neighbourhoods.
f. Relevant climate and atmospheric conditions, precipitation, evaporation, wind
direction and frequency of occurrence, temperature and seasonal variability.
g. Current and future settlement areas.
h. Flora and fauna description in project area and neighbourhoods, existing habitats
or plant communities, location of sensitive or rare species, protected sites.

15. 6. Mitigation measures
a. Control and reducing measures for discharge of untreated wastewater into river.
b. Use of low quality water for sprinkling for dust prevention on working sites.
c. Regular inspection of the sewage collecting system in order to timely detect any
failures, and take proper action.
d. Topsoil removal and storage in separate piles and reinstallation after refilling of
trenches, to enable natural vegetation.
e. Reduction of overall harmful emissions.
f. Planting of vegetation on borders of STP sites.
g. Covered treatment basins or covered structures for the sludge treatment and
storage.
h. Insulating pump house
i. Use of low noise and vibration installations and equipment.
j. Limiting animal access to the locations that might consist a risk.
k. Provision of new appropriate habitat.
l. Creating opportunities for fauna migration.

16. 7. Terms of Reference (TOR) or guidelines
These are the summary of the findings of the scoping activities. Specific guidelines
should include all key aggrements reached during the scoping period on issues and
alternatives.

17. 3. A wastewater treatment plant has to process a peak flow of 40000m3/d. Design criteria
for surface overflow rates have been set by the state regulatory agency at a maximum of
100m/d. Design the primary clarifier if it is a long rectangular basin. Design the
completely mix activated sludge plant if the wastewater has a BOD5 of 1200mg/l that
must be reduced to 200mg/l prior to discharge to a municipal sewer. Pilot plant analysis
indicates that a mean cell residence time of 5 day maintaining MLSS concentration of
5000mg/l produces the desired result. The value of Y is determined to be 0.07kg/kg and
the value of kd is found to be 0.03d-1.
Solution:-
a. Design of primary clarifier
Assuming a peak factor of 2.5, the average flow rate to the WWTP can be calculated by
the following expression-
Qavg = Qpeak/ peaking factor
= 40000(m3/d) / 2.5
= 16000(m3/d).
Assuming average surface overflow rate of 40m/d, the required surface area of the long
rectangular basin can be calculated by the following expression-
As = Qavg / SLR
= 16000(m3/d) / 40(m/d)
= 400m2.
Providing two channels , surface area of each channel = 200m2

18. Assuming a length : width ratio of 3:1,
L*W = 200
or, 3W*W = 200
Thus, W = 8.16m
L = 24.5m
Assuming a detention time of 2h,
Volume of each channel = 8000(m3/d) * (2/24)(d)
= 666.66m3
Computing the side water depth or liquid depth of the tank,
D = V(volume of the tank)/ As (surface area of the tank)
= 666.66(m3)/200(m2)
= 3.3m
Assuming a 60% removal of suspended solids on dry weight basis in primary clarifier, the
initial BOD5 is calculated to be 3000mg/L [1200(mg/L)/0.40]
Thus, the mass of primary sludge produced in each tank is calculated as follows:
Msl =60% of suspended solids in the effluent
= 0.60 * 3000(mg/L) * 103(L/m3) * 8000(m3/d)
= 14400 kg/d
= 600 kg/h
Assuming the specific gravity of primary sludge as 1.03 and 6% of solids content in sludge,
the volume of the sludge produced each day can be computed by:

19. Vsl = Msl / (ƿw * ssl * Ps)
where, Vsl = volume of sludge, m3/d
Msl = mass of sludge, kg/d
ƿw = density of water, kg/m3
ssl = specific gravity of primary sludge
Ps = percentage of solids in primary sludge expressed as decimal
Thus, Vsl = 14400(kg/d) /( 998.2(kg/m3) * 1.03 * 0.06)
= 233 (m3/d)
= 9.70 (m3/h)
Assuming sludge is removed at every four hours by pumping from the trapezoidal hopper
bottom, the capacity of the hopper bottom required will be equal to the sludge collected
every 4 hours.
Capacity of sludge pocket, C= 9.70(m3/h) * 4(h)
= 38.8 m3
Assuming A = 5m, B = 6m and H = 1.3m, the volume can be calculated by
V = 1/3 H (B2 + AB + B2)
Substituting, V = 1/3 * 1.3 * (42 + 4*5 + 52)
= 39.43 m3.
Assuming 10% slope of the tank bottom, the depth for slope will be
= (24.5 – 6)*0.1
= 1.85m

20. So, the overall depth of the tank can be worked out as
D = liquid depth + freeboard + bottom slope depth + hopper bottom depth
= 3.3 m + 0.3 m (assuming) + 1.85 m + 1.3 m
= 6.75 m
Overall length of the tank providing 10% length for inlet and outlet zones
L = 24.5 + 2.45 = 26.95m = 27m
Overall width of the tank with 2 channels and including 3 walls each of 0.2 m thickness, is
= 2*8.16 + 3*0.2 = 16.92m = 17m
Check for surface overflow rate at peak flow
Surface overflow rate = Qpeak / Surface area = (40000(m3/d)/2)/ 200m2 = 100 m/d
( Acceptable as the permissible SOR at Qpeak = 100 m/d)
Design summary:
1. No. of units = 2
2. Volume of each channel = 667 m3
3. Overall length of tank = 27m
4. Overall width of tank = 17m
5. Overall depth of tank = 6.75m
6. Hopper bottom capacity (for 4 hours sludge accumulation) = 38.8m3

21. Sectional view of primary clarifier
Top view of primary clarifier

22. b. Design of a completely mixed activated sludge plant
Computing the reactor volume,
V= (Qavg * Y * Ɵc * (S0 – S)) / (X(1 + kd Ɵc))
where,
Y = cell yield coefficient
Ɵc = mean cell residence time, (d)
S0 = influent substrate concentration, (mg/l)
S = substrate concentration in effluent, (mg/l)
X = biomass concentration, (mg/l)
kd = endogeneous decay coefficient (d-1)
Thus, volume of the reactor
= 16000(m3/d) * 0.7(kg/kg) *5(d) * (1200-200)(mg/l)
5000(mg/l) * ( 1+ 0.03(d-1)*5(d))
= 9739.13 m3.
Computing the hydraulic retention time (HRT), Ɵ
HRT, Ɵ = V/Q = 9739.13/16000
= 0.608(d)
= 14.60 h
The mass of sludge wasted can be computed from the following relation,
Qw * XR = (V * X)/ Ɵc
= (9739.13m3 * (5000 * 10-3)(kg/m3) )/5(d) = 9739.13 kg/d

23. Computing oxygen required for aeration from the relation,
O2 required per day = (mass of BOD5 utilized) – 1.42* (mass of biomass wasted)
= {Qavg* (S0-S)} – 1.42 *(Qw* XR)
=[16000(m3/d) * {(1200-200) * 10-3(kg/m3)}]-[1.42*9739.13(kg/d)]
=2170.4354 kg/d
Theoretical amount of air needed for aeration,
Qair = (mass of O2 required)/(air density * % of O2 in air)
= 2170.4354 (kg/d) / (1.192 (kg/m3) * 0.232)
= 7848.42 m3/d
Computing the recirculation ratio, R , assuming Xe = 0 and XR = 10000mg/l
R= [(Qavg *X) - (Qw* XR)] / [Qavg(XR – X)]
=[ {16000(m3/d) * 5 (kg/m3)} - 9739.13 kg/d] / [16000(m3/d) * {10-5}(kg/m3)]
= 0.878
Thus, flow to be recycled QR = R*Q = 0.878 * 16000 (m3/d) = 14052.174 m3/d
and, flow to be wasted Qw = (Qw * XR)/ XR = 973.913 m3/d
c. Design of secondary clarifier
Calculating the surface area of the tank assuming SOR to be 15m/d
As = Q/SOR
= 16000 (m3/d) / 15 (m) = 1066.66 m2
Providing two tanks, Surface area of each tank = 533.33 m2
Computing diameter of the tank, d = (4*As/∏)1/2 = 26 m

24. Assuming SWD, D = 3.7 m, the effective volume of the tank
V = As * D
= 533.33 m2 * 3.7 m
= 1973.31 m3
Providing a freeboard of 0.3 m, the total depth of the tank, D = 4 m
Total volume of the tank = 2133.33 m3
Checks for each tank are as follows:
For surface loading rate at peak flow,
SLR = peak flow/surface area =[ 20000 (m3/d) /533.33 (m2)]
= 37 (m3/m2d)
Acceptable as lower than the range of 40-64 (m3/m2d)
For weir loading rate at peak flow,
WLR = peak flow/weir length = [20000 (m3/d) / ∏*26 (m)] = 245 (m3/m-d)
Hydraulic loading rate, HRT, Ɵ = effective volume of tank/ flow of wastewater
= 1973.31 (m3)/ 20000 (m3/d)
= 0.099 (d)
= 2.36 h
Acceptable.

25. Design summary of reactor:
1. Q0 = 16000 mg/l
2. S0 = 1200 mg/l
3. X0 = 5000 mg/l
4. S = 200 mg/l
5. V = 9739.13 m3
6. QR = 14052.174 mg/l
7. Qw = 973.913 m3/d
8. XR = 10000mg/l
Design summary of secondary clarifier
1. No of tanks = 2
2. Effective volume of each tank = 1973.31 m3
3. Total volume of each tank = 2133.33 m3
4. Diameter of each tank = 26 m
5. Side water depth in each tank = 3.7 m
6. Freeboard = 0.3 m
7. Total depth = 4 m
8. Hydraulic retention time = 2.36 h

26. THANK YOU