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
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
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
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
