The document provides solutions to various problems related to environmental engineering. It includes: 1) Calculating the area required for landfilling garbage over 5 years for a city with a population of 2.38 million, accounting for waste generation growth and the percentage reaching the landfill. The required area is estimated to be 138-143 hectares depending on the soil content in the landfill. 2) Drawing a diagram showing the typical components of a sewage treatment plant, including primary and secondary treatment units, and designing the secondary treatment unit using a complete mix activated sludge process for a city with a population of 200,000. 3) Estimating the moisture content, heating value, and chemical composition of the

1. by
Ph.D. Qualifying Examination
Presentation
Rajat Pundlik
S19CE09001
Environmental Engineering
School of Infrastructure
Indian Institute of Technology Bhubaneswar
November, 2019

2. 1. (a) A certain stimulus administrated to each of the 12 patient resulted in
the following increase in blood pressure:
5, 2, 8, -1, 3, 0, -2, 1, 5, 0, 4, and, 6
Can it be concluded that the stimulus will, in general, be accompanied by
an increase in blood pressure?
Given, critical value for testing hypothesis is 1.80.
2

3. Solution : Null hypothesis, H0 : µx = µy , i.e.,
there is no significant increase in blood pressure
due to stimulus
Alternative Hypothesis, H1 : µx < µy ,
there is increase in blood pressure due to
stimulus
3
d d2
5 25
2 4
8 64
-1 1
3 9
0 0
-2 4
1 1
5 25
0 0
4 16
6 36
Σd = 31 Σd2 = 185

4. Solution : t –test statistic : t =
𝑑̅
𝑆/⎷𝑛
S2 =
1
𝑛−1
[Σd2 -
Σd 2
𝑛
]
= 9.5382
𝑑̅ =
Σd
𝑛
= 2.58
t =
𝑑̅
𝑆/⎷𝑛
=
2.58
9.5382/⎷12
= 2.89
Since calculated t > 1.80, H0 is rejected
Hence, there is increase in blood pressure due to stimulus.
4

5. 1. (b) The data about the sales and advertisement expenditure of a firm is given below:
Coefficient of correlation is 0.9.
(i) Estimate the likely sales for a proposed advertisement expenditure of INR 10 crores.
(ii) What should be the advertisement expenditure if the firm proposes a sales target of
INR 60 crores.
5
Sales (in crores of INR) Advertisement
expenditure (in crores
of INR)
Mean 40 6
S.D. 10 1.5

6. Solution : Given Value:
Sales (in crores of INR) mean, 𝑥′ = 40
Advertisement expenditure (in crores of INR) mean, 𝑦′ = 6
Sales (in crores of INR) s.d. , σ𝑥 = 10
Advertisement expenditure (in crores of INR) s.d. , σ𝑦 = 1.5
Coefficient of correlation, r = 0.9
For Regression Equation
( y – y’ ) = r * σ𝑦
σ𝑥
( x – x’ )
( y – 6 ) = 0.9 * 1.5
10
( x – 40 )
6

7. 7
(a). Likely sales for a proposed advertisement expenditure of INR 10 crores
( 10 – 6 ) = 0.9 * 1.5
10
( x – 40 )
x = 69.63 (in crores of INR)
(b). Advertisement expenditure if the firm proposes a sales target of INR 60
crores
( y – 6 ) = 0.9 * 40
10
( 60 – 40 )
y = 8.7 (in crores of INR)

8. 1. (c) The number of scooter accidents per month in a certain town were as
follows:
12, 8, 20, 2, 14, 10, 15, 6, 9, 4.
Are these frequencies in agreement with the belief that accident
conditions were the same during this 10 month period?
Critical value: 16.919
8

9. 9
H0: There is no significant difference in accident condition.
H1: There is significant difference in accident condition

10. 10
χ2 = 26.6; df = n-1=10-1 = 9
χ2(9) = 16.919
χ2 > χ2(9) H0 is rejected
Hence, Accident condition were not same.

11. 1. (a)Estimate the moisture content and chemical composition of the
organic fraction of the MSW, with the properties shown below
Estimate the heating value of individual material and the mixed
solid waste using modified Dulong Equation
HV (
𝑘𝐽
𝑘𝑔
) = 337 [C] + 1419 [H2 – 0.125 O2] + 93 [S] + 23 [N]
Where, C, H, O, N and S are the composition on dry basis. 11
Description
Weight (kg) Composition (kg)
Wet Dry C H O N S Ash
Paper 19 16 6.96 0.96 7.04 0.048 0.032 0.96
Plastic 3.7 3.5 2.1 0.252 0.798 0 0 0.35
Food wastes 5.1 1.90 0.912 0.122 0.714 0.0494 0.0076 0.095
Yard wastes 8.4 2.6 1.243 0.156 0.988 0.0884 0.0078 0.117
Textiles 1 0.8 0.44 0.053 0.25 0.0368 0.0012 0.02
Rubber 0.22 0.2 0.172 0.022 0 0.0044 0 0.022
Wood 1.3 0.9 0.446 0.054 0.384 0.0018 0.0009 0.0135

12. 12
Description
Weight (kg)
Wet Dry
Paper 19.0000 16.00000
Plastic 3.70000 3.50000
Food wastes 5.10000 1.90000
Yard wastes 8.40000 2.60000
Textiles 1.00000 0.80000
Rubber 0.22000 0.20000
Wood 1.30000 0.90000
38.720 25.900
Moisture content = (
𝑊𝑒𝑡 𝑤𝑒𝑖𝑔ℎ𝑡 −𝐷𝑟𝑦 𝑤𝑒𝑖𝑔ℎ𝑡
𝑊𝑒𝑡 𝑤𝑒𝑖𝑔ℎ𝑡
∗ 100)
= (
38.720 −25.900
38.720
∗ 100 )
= 33.11%

13. 13
Description
Weight (kg) Composition (kg)
Wet Dry C H O N S Ash
Paper 19.0000 16.00000 6.96000 0.96000 7.04000 0.04800 0.03200 0.96000 16.000
Plastic 3.70000 3.50000 2.10000 0.25200 0.79800 0.00000 0.00000 0.35000 3.5000
Food wastes 5.10000 1.90000 0.91200 0.12200 0.71400 0.04940 0.00760 0.09500 1.9000
Yard wastes 8.40000 2.60000 1.24300 0.15600 0.98800 0.08840 0.00780 0.11700 2.6002
Textiles 1.00000 0.80000 0.44000 0.05300 0.25000 0.03680 0.00120 0.02000 0.8010
Rubber 0.22000 0.20000 0.17200 0.02200 0.00000 0.00440 0.00000 0.02200 0.2204
Wood 1.30000 0.90000 0.44600 0.05400 0.38400 0.00180 0.00090 0.01350 0.9002
38.720 25.900 12.273 1.619 10.174 0.2288 0.0495 1.5775

14. 14
Wt. (kg)
Molecular Wt.
(kg/kmoles)
MOLES
NORMALISED
WITH SULPHUR
C 12.27300 12 1.02275 660
H 1.61900 1 1.619 1045
O 10.17400 16 0.635875 410
N 0.22880 14 0.016343 11
S 0.04950 32 0.001547 1
Chemical Composition of the organic fraction :
C660 H1045 O410 N11 S

15. 15
Description
Composition in %
C H O N S ASH
Paper 43.5 6 44 0.3 0.2 6
Plastic 60 7.2 22.8 0 0 10
Food wastes 48 6.42 37.58 2.6 0.4 5
Yard wastes 47.81 6 38 3.4 0.3 4.5
Textiles 55 6.63 31.25 4.6 0.15 2.5
Rubber 86 11 0 2.2 0 11
Wood 49.56 6 42.67 0.2 0.1 1.5
Heating value of individual material using modified Dulong Equation
HV (
𝑘𝐽
𝑘𝑔
) = 337 [C] + 1419 [H2 – 0.125 O2] + 93 [S] + 23 [N]
Description
Composition in %
C H O N S ASH
Paper 43.5 6 44 0.3 0.2 6
Plastic 60 7.2 22.8 0 0 10
Food wastes 48 6.42 37.58 2.6 0.4 5
Yard wastes 47.81 6 38 3.4 0.3 4.5
Textiles 55 6.63 31.25 4.6 0.15 2.5
Rubber 86 11 0 2.2 0 11
Wood 49.56 6 42.67 0.2 0.1 1.5
Mixed solid
waste
47.39 6.25 39.28 0.88 0.19 6.09

16. 16
Description Heating value (
𝑘𝐽
𝑘𝑔
)
PAPER 15394.5
PLASTIC 26392.65
FOOD WASTE 18718.90789
YARD WASTE 17991.04231
TEXTILES 22512.65625
RUBBER 44641.6
WOOD 17660.12222
MIXED SOLID WASTE 17909.70907

17. 1. (b) The average solid waste generation in an Indian city is 1.15 kg/capita/d for four
months and 0.8 kg/capita/day for eight months, in a year. The population of the city is
2.38 million as per the recent census. It is planned to earmark land for landfilling
garbage for a period of five years. Assuming the landfill garbage density of 450 kg/m3, 10
m lift and the percentage of garbage reaching the landfill facility as 85% of the garbage
generated in the city, determine the landfill area for (a) 25% soil in cell volume (b) 30%
soil in cell volume. Assume any other data if necessary.
17

18. Solution :
1. Current waste generation per year
1. The average solid waste generation
for first four month is 1.15 kg/capita/d and
for next eight month of year 0.8 kg/capita/day
2. Population of city is 2.38 million = 23, 80, 000
3. Waste generation
for first four month = 1.15 * 23, 80,000 * 4 * 30 = 328,440,000 kg
= 328,440 tons
for next eight month = 0.8 * 23, 80,000 * 8 *30 = 456,960,000 kg/d
= 456,960 tons/d
Therefore, Current waste generation per year = 785400 tons per year
18

19. 2. Estimated rate of increase of waste generation per year = 8.3 percent
(Using rate of population growth where waste generation growth rate estimates not
available)
3. Proposed life of landfill = 5 years
4. Waste generation after n years = 785400 ( 1 + 8.3/100 )5 =1170127.484 tons per year
5. Total waste generation in n years (T) in tons =
1
2
∗ 785400 + 1170127.484 ∗ 5
= 4,888,818.709 (tons)
6. Total waste reaching Landfill = 0.85 * 4,888,818.709
= 4,155,495.903 (tons)
7. Total volume of waste in 5 years (Vw) = 4,155,495.903 / 0.45 (t/cum)
= 9,234,435.339 cum
8. Total volume of daily cover in 5 years (Vdc)
for 25% soil in cell volume = 0.25 * Vw
= 2,308,608.835
for 30% soil in cell volume = 0.25 * Vw
= 2,770,330.602 19

20. 8. Total volume required for components of liner system and cover system
Vc = 0.25 * Vw
= 2,308,608.835 cum
9. Landfill capacity (Ci)
Ci = Vw + Vd + Vc
for 25% soil in cell volume, Ci = 13,851,653.01 cum
for 25% soil in cell volume, Ci = 14,313,374.78 cum
10. Landfill Height and Area
Landfill height (Hi)= 10 m (Given)
Area required for landfill (Ai) = Ci / Hi
for 25% soil in cell volume, Ai = 1,385,165.301 sqm = 138.516 hectare
for 30% soil in cell volume, Ai = 1,431,337.478 sqm = 143.134 hectare
Total area required (including infrastructural facilities)
Aif = 1.15 Ai
for 25% soil in cell volume, Aif = 1,592,940.096 sqm = 159.294 hectare
for 30% soil in cell volume, Aif = 1,646,038.099 sqm = 164.604 hectare 20

21. 1. A wastewater treatment plant has been proposed for the city A. Current
population of the city is 2 Lakhs. Draw the STP showing different
components and design the secondary treatment unit considering
complete mix ASP process. Assume other necessary data.
21

22. Sewage treatment plant with different componenets
22
Bar
screen
Grit
Chamber
Parshall
flume
Aeration
tank
PST
Influent
Wastewater
SST
AD
Air
Supernatant
Recycled sludge
Secondary
sludge
Primary
sludge
Gases
Effluent
for further
treatment
or disposal
Digested sludge
Sludge
handling &
disposal
PST: Primary Sedimentation Tank
SST: Secondary Sedimentation Tank
AD: Anaerobic Digester

23. Design of aeration basin considering ASP
Given, Population = 2 lacks = 200000 people
Assumptions:
23

24. (a) Estimation of wastewater flow to the ASP, Q
𝑄 = 𝑃𝑜𝑝𝑢𝑙𝑎𝑡𝑖𝑜𝑛 × 𝑑𝑎𝑖𝑙𝑦 𝑤𝑎𝑡𝑒𝑟 𝑠𝑢𝑝𝑝𝑙𝑦 × 0.8
= 200000 𝑝𝑒𝑜𝑝𝑙𝑒 × 135 𝐿 𝑐𝑎𝑝𝑖𝑡𝑎 − 𝑑 × 0.8 × 10−3
𝑚3
/𝐿
= 21600 𝑚3/𝑑
(b) Estimation of treatment efficiency, E
𝐸 =
𝑆0 − 𝑆
𝑆0
× 100
=
200 − 30 𝑚𝑔/𝐿
200 𝑚𝑔/𝐿
× 100
= 85%
(c) Computation of the reactor volume, V
1
θ𝑐
=
𝑌 × (𝑆0 − 𝑆)
θ × 𝑋
− 𝐾𝑑̅ [𝑤ℎ𝑒𝑟𝑒, θ =
𝑉
𝑄
]
𝑉 =
𝑄 × 𝑌 × θ𝑐 × (𝑆0 − 𝑆)
𝑋 × (1 + 𝐾𝑑̅θ𝑐)
=
21600 𝑚3
𝑑 × 0.5 × 10 𝑑 × 200 − 30 𝑚𝑔/𝐿
3500 𝑚𝑔 𝐿 × [1 + 0.06 𝑑−1 × 10 𝑑 ]
= 3278.57 𝑚3
24

25.  Biomass Mass Balance:
As steady state condition prevails, accumulatio = 0
𝑄0𝑋0 + 𝑉
𝑑𝑋
𝑑𝑡 𝑛𝑒𝑡
= 𝑄0 − 𝑄𝑤 𝑋𝑒 + 𝑄𝑤𝑋𝑟
Assuming X0 and Xe = 0,
𝑉
𝑑𝑋
𝑑𝑡 𝑛𝑒𝑡
= 𝑄𝑤𝑋𝑟
𝑉
µ𝑚𝑆𝑋
𝐾𝑠 + 𝑆
− 𝐾𝑑̅𝑋 = 𝑄𝑤𝑋𝑟
µ𝑚𝑆
𝐾𝑠 + 𝑆
=
𝑄𝑤𝑋𝑟
𝑉𝑋
+ 𝐾𝑑̅ (1)
Accumulation
of biomass
Inflow of
biomass
Net growth
of biomass
Outflow of
biomass
= +
̶
Influent
biomass
+ Biomass
production
= Effluent
biomass
+ Wasted
biomass
V, X, SS
Q0 , X0 Q0 + Qr Qe , Xe
Qr , Xr
Qw , Xw
S0 X, S Se
25

26.  Substrate Mass Balance:
𝑄0𝑆0 + 𝑉
𝑑𝑆
𝑑𝑡
= 𝑄0 − 𝑄𝑤 𝑆 + 𝑄𝑤𝑆
𝑄0𝑆0 −
𝑉
𝑌
µ𝑚𝑆𝑋
𝐾𝑠 + 𝑆
= 𝑄0𝑆
µ𝑚𝑆
𝐾𝑠 + 𝑆
=
𝑌𝑄0 𝑆0 − 𝑆
𝑉𝑋
2
Equating Eq. (1) and (2),
𝑄𝑤𝑋𝑟
𝑉𝑋
+ 𝐾𝑑̅ =
𝑌𝑄0 𝑆0 − 𝑆
𝑉𝑋
Now, θ =
𝑉
𝑄0
and θ𝑐 =
𝑉𝑋
𝑄𝑤𝑋𝑟+𝑄𝑒𝑋𝑒
where Xe=0, θ𝑐 =
𝑉𝑋
𝑄𝑤𝑋𝑟
Substituting these in Eq. (3) we get,
V, X, SS
Q0 , X0 Q0 + Qr Qe , Xe
Qr , Xr
Qw , Xw
S0 X, S Se
Inflow of
substrate
+ Consumption
of substrate
= Outflow of
substrate
+ Substrate
wasted
(3)
1
θ𝑐
+ 𝐾𝑑̅ =
𝑌(𝑆0 − 𝑆)
θ𝑋
26

27. (d) Determination of hydraulic retention time, θ
θ = 𝑉/𝑄
= 3278.57 𝑚3
/21600 𝑚3 𝑑
= 0.15 𝑑 = 3.64 ℎ ~ 4 ℎ
Determination of volume based on θ = 4 h
𝑉 = 𝑄 × θ
= 21600 𝑚3
/𝑑 ×
4
24
𝑑
= 3600 𝑚3
(e) Determination of sludge wasting rate, Qw
θ𝑐 =
𝑉𝑋
𝑄𝑤𝑋𝑤 + 𝑄𝑒𝑋𝑒
Neglecting effluent biomass concentration, Xe = 0
𝑄𝑤 =
𝑉𝑋
θ𝑐𝑋𝑤
=
3600 𝑚3
× 3500 𝑚𝑔 𝑀𝐿𝑉𝑆𝑆/𝐿
10 𝑑 × 0.8 × 10000 𝑚𝑔 𝑀𝐿𝑉𝑆𝑆/𝐿
= 126 𝑚3
𝑑
Q , X0 Q + Qr Qe , Xe
Qr , Xr
Qw , Xw
27

28. (f) Mass of sludge wasted per day from the recycle line,
𝑄𝑤 × 𝑋𝑟 = 126 𝑚3
𝑑 × 0.8 × 10000 𝑚𝑔 𝑀𝐿𝑉𝑆𝑆/𝐿 × 10−6
𝑘𝑔 𝑚𝑔 × 103
𝐿 𝑚3
= 1008 𝑘𝑔/𝑑
(g) Determination of Oxygen requirement for aeration,
Mass of ultimate BOD utilized per day,
𝐵𝑂𝐷𝐿 =
𝑄 × (𝑆0 − 𝑆)
𝑓
[𝑤ℎ𝑒𝑟𝑒, 𝑓 = 0.68]
=
21600 𝑚3
𝑑 × 200 − 30 𝑚𝑔/𝐿 × 10−6
𝑘𝑔 𝑚𝑔 × 103
𝐿 𝑚3
0.68
= 5400 𝑘𝑔/𝑑
𝑂𝑥𝑦𝑔𝑒𝑛 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑚𝑒𝑛𝑡 = 𝑚𝑎𝑠𝑠 𝑜𝑓 𝐵𝑂𝐷𝐿 𝑢𝑡𝑖𝑙𝑖𝑧𝑒𝑑
− 1.42 × (𝑚𝑎𝑠𝑠 𝑜𝑓 𝑏𝑖𝑜𝑚𝑎𝑠𝑠 𝑤𝑎𝑠𝑡𝑒𝑑)
= 5400 𝑘𝑔/𝑑 − 1.42 × 1008 𝑘𝑔/𝑑
= 3968.64 𝑘𝑔 𝑑 = 165.36 𝑘𝑔/ℎ
(h) Estimation of quantity of air required for aeration, Qair
𝑄𝑎𝑖𝑟 =
𝑚𝑎𝑠𝑠 𝑜𝑓 𝑂2 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑑
𝑎𝑖𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 × % 𝑂2 𝑖𝑛 𝑎𝑖𝑟
[𝐴𝑖𝑟 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 = 1.192 𝑘𝑔 𝑚3
𝑎𝑡 23℃]
=
3968.64 𝑘𝑔 𝑑
1.192 𝑘𝑔 𝑚3 × 0.23
= 14475.63 𝑚3
𝑑 = 603 𝑚3
ℎ
28

29. (i) Determination of recirculation ratio, R and flow to be recycled, Qr
𝑅 =
𝑄𝑟
𝑄
=
𝑋
𝑋𝑟 − 𝑋
=
3500 𝑚𝑔/𝐿
0.8 × 10000 − 3500 𝑚𝑔/𝐿
= 0.78
[Range: 0.25 – 1]
(j) Computation of food to microorganism ratio, F/M
𝐹
𝑀
=
𝑄0 × 𝑆0
𝑉 × 𝑋
=
21600 𝑚3
𝑑 × 200 𝑚𝑔 𝐿
3600 𝑚3 × 3500 𝑚𝑔/𝐿
= 0.34 𝑑−1
[Range: 0.2 – 0.4 d-1]
(k) Computation of volumetric loading rate,
𝑄 × 𝑆0
𝑉
=
21600 𝑚3 𝑑 × 200 𝑚𝑔 𝐿 × 10−6 𝑘𝑔 𝑚𝑔 × 103 𝐿 𝑚3
3600 𝑚3
= 1.2 𝑘𝑔 𝐵𝑂𝐷5 𝑚3. 𝑑 𝑜𝑓 𝑟𝑒𝑎𝑐𝑡𝑜𝑟 𝑣𝑜𝑙𝑢𝑚𝑒
[Range: 0.8 – 2 kg BOD5/m3 .d]
29

30. Design of secondary sedimentation tank
(a) Computation of the surface area of the tank,
Assuming, surface overflow rate (SOR) 15 m3/m2-d the surface area of the tank, As
𝐴𝑠=
𝑄
𝑆𝑂𝑅
=
21600 𝑚3
𝑑
15 𝑚3 𝑚2. 𝑑
= 1440 𝑚2
Providing two tanks of surface area of each tank equal to 720 m2.
(b) Determination of the diameter of each tank, d
𝑑 =
4 × 𝐴𝑠
π
=
4 × 720
π
= 30.277𝑚 ~ 30𝑚
Providing the tank of 30m diameter the surface area of each tank will be.
=
π
4
× 302
= 706.85 𝑚2
~ 707 𝑚2
(c) Computation of total volume of each tank, VSST
Assuming, side water depth, D = 3.7 m, the effective volume of each tank, V
𝑉 = 𝐴𝑠 × 𝐷 = 707 𝑚2
× 3.7 𝑚 = 2615.9 𝑚3
Providing free board of 0.3 m, the total depth of each tank, 3.7 + 0.3 = 4 m
The total volume of each tank, VSST
𝑉𝑆𝑆𝑇 = 707 𝑚2
× 4 𝑚 = 2828 𝑚3 30

31. Checks for each tank:
(a) Surface loading rate at peak flow, SLR
𝑆𝐿𝑅 =
𝑝𝑒𝑎𝑘 𝑓𝑙𝑜𝑤
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
=
2.5 × 10800 𝑚3/𝑑
707 𝑚2
[𝑄𝑆𝑆𝑇 =
𝑄
2
=
21600 𝑚3
𝑑
2
]
= 38.19 𝑚3
𝑚2
. 𝑑
[Range: 40 – 64 m3/m2.d]
(b) Weir loading rate at peak flow, WLR
𝑊𝐿𝑅 =
𝑝𝑒𝑎𝑘 𝑓𝑙𝑜𝑤
𝑤𝑒𝑖𝑟 𝑙𝑒𝑛𝑔𝑡ℎ
=
2.5 × 10800 𝑚3/𝑑
π × 30 𝑚
= 286.48 𝑚3
𝑚. 𝑑
[Range: 250 – 375 m3/m.d]
(c) Solids loading rate at peak flow, SL
𝑆𝐿 =
𝑡𝑜𝑡𝑎𝑙 𝑠𝑜𝑙𝑖𝑑𝑠
𝑠𝑢𝑟𝑓𝑎𝑐𝑒 𝑎𝑟𝑒𝑎
=
2.5 × 10800 𝑚3 𝑑 × 3500 𝑔 𝑚3
707 𝑚2
= 133.6 𝑘𝑔 𝑚2 . 𝑑
[Range: less than 192 kg/m2.d ]
(d) Hydraulic retention time, θ
θ =
𝑒𝑓𝑓𝑒𝑐𝑡𝑖𝑣𝑒 𝑣𝑜𝑙𝑢𝑚𝑒 𝑜𝑓 𝑡𝑎𝑛𝑘
𝑓𝑙𝑜𝑤 𝑜𝑓 𝑤𝑎𝑠𝑡𝑒𝑤𝑎𝑡𝑒𝑟
=
2828 𝑚3
10800 𝑚3 𝑑
= 0.26 𝑑 ≅ 6.28 ℎ
31

32. 32
Thank
You