Question 1 |
Which of the following statements is/are TRUE for the aerobic composting of sewage sludge?
Bulking agent is added during the composting process to reduce the porosity of the solid mixture | |
Leachate can be generated during composting | |
Actinomycetes are involved in the process | |
In-vessel composting systems cannot be operated in the plug-flow mode |
Question 1 Explanation:
A bulking agent (wood ships shredded leaves or types) is intended to support the structure of sludge by increasing its porosity to encourage effective aeration.
In-vessel composting systems can be divided into plug flow and agitated bed modes. In plug flow mode, the relationship between particles in the composting mass stays the same through the process and the system operates on first-in, first-out principle. In an agitated bed system, composting material is mixed mechanically during the processing.
The heat produced in aquatic composting cause any surplus moisture to be extracted as water vapour and hence no leachate is produced.
Actinomycetes are aerobic, spore forming positive bacteria. These are thermophilic species and their development depends on aerobic conditions, temperature and water content. The heat produced in aerobic composting helps support the growth of these species.
Leachate can generated from vegetable wastes.
In-vessel composting systems can be divided into plug flow and agitated bed modes. In plug flow mode, the relationship between particles in the composting mass stays the same through the process and the system operates on first-in, first-out principle. In an agitated bed system, composting material is mixed mechanically during the processing.
The heat produced in aquatic composting cause any surplus moisture to be extracted as water vapour and hence no leachate is produced.
Actinomycetes are aerobic, spore forming positive bacteria. These are thermophilic species and their development depends on aerobic conditions, temperature and water content. The heat produced in aerobic composting helps support the growth of these species.
Leachate can generated from vegetable wastes.
Question 2 |
Which one of the following statements is TRUE for Greenhouse Gas (CHG) in the atmosphere?
GHG absorbs the incoming short wavelength
solar radiation to the earth surface, and allows
the long wavelength radiation coming from the
earth surface to pass through. | |
GHG allows the incoming long wavelength solar
radiation to pass through to the earth surface,
and absorbs the short wavelength radiation coming from the earth surface. | |
GHG allows the incoming long wavelength solar
radiation to pass through to the earth surface,
and allows the short wavelength of radiation
coming from the earh surface to pass through | |
GHG allows the incoming short wavelength
solar radiation to pass through to the earth
surface, and absorbs the long wavelength
radiation coming from the earth. |
Question 2 Explanation:
Green House Gas allow short wave radiation to pass through to the earth surface and absorbs the long wavelength radiation coming from the earth surface.
Question 3 |
A sewage treatment plant receives sewage at a flow rate of 5000 m^3/day. The
total suspended solids (TSS) concentration in the sewage at the inlet of primary
clarifier is 200 mg/L. After the primary treatment, the TSS concentration in
sewage is reduced by 60%. The sludge from the primary clarifier contains 2%
solids concentration. Subsequently, the sludge is subjected to gravity thickening
process to achieve a solids concentration of 6%. Assume that the density of
sludge, before and after thickening, is 1000 kg/m^3.
The daily volume of the thickened sludge (in m^3/day) will be_________. (round off to the nearest integer)
The daily volume of the thickened sludge (in m^3/day) will be_________. (round off to the nearest integer)
5 | |
10 | |
20 | |
30 |
Question 3 Explanation:
Wt. of TSS at inlet of PST
=5000 \times 10^3\frac{L}{d} \times 200 \times 10^{-6}Kg/L=1000Kg/d
Wt. of solid in sludge from PST =0.6 \times 1000=600 Kg/d
Wt. of sludge before thikeninig =\frac{600}{\frac{2}{100}}=30000 Kg/d
Wt. of sludge before thikeninig =\frac{600}{\frac{6}{100}}=10000 Kg/d
Daily volume of thickened sludge =frac{10000}{1000}=10m^3
=5000 \times 10^3\frac{L}{d} \times 200 \times 10^{-6}Kg/L=1000Kg/d
Wt. of solid in sludge from PST =0.6 \times 1000=600 Kg/d
Wt. of sludge before thikeninig =\frac{600}{\frac{2}{100}}=30000 Kg/d
Wt. of sludge before thikeninig =\frac{600}{\frac{6}{100}}=10000 Kg/d
Daily volume of thickened sludge =frac{10000}{1000}=10m^3
Question 4 |
A city generates 40 \times 10^{6} kg of municipal solid waste (MSW) per year, out of which only 10% is recovered/recycled and the rest goes to landfill. The landfill has a single lift of 3 m height and is compacted to a density of 550 kg/m^{3}. If 80% of the landfill is assumed to be MSW, the landfill area (in m^{2}, up to one decimal place) required would be ______
25743.3 | |
36548.4 | |
27272.7 | |
26845.6 |
Question 4 Explanation:
Total weight generated by city
=40\times 10^{6}\: kg/year
Weight of MSW going into landfill
=0.9\times 40\times 10^{6}\: kg/year
=36\times 10^{6}\: kg/year
Compacted density
=550\: kg/m^{3}
Compacted volume of MSW
=\frac{36\times 10^{6}\: kg/year}{550\: kg/m^{3}}
=65454.5454\: m^{3}/year
Total landfill volume = Vol. of MSD + Vol. of cover
Given, Volume of MSW =0.8\times Total landfill volume
\therefore Vol. of cover =0.2\times Total landfill volume
\therefore Total landfill volume =\frac{65454.5454}{0.8}m^{3}/year =81818.18175m^{3}/year
Height of landfill =3\: m
\therefore Area of lanfill=\frac{81818.18175}{3} =27272.7\: m^{2}/year
=40\times 10^{6}\: kg/year
Weight of MSW going into landfill
=0.9\times 40\times 10^{6}\: kg/year
=36\times 10^{6}\: kg/year
Compacted density
=550\: kg/m^{3}
Compacted volume of MSW
=\frac{36\times 10^{6}\: kg/year}{550\: kg/m^{3}}
=65454.5454\: m^{3}/year
Total landfill volume = Vol. of MSD + Vol. of cover
Given, Volume of MSW =0.8\times Total landfill volume
\therefore Vol. of cover =0.2\times Total landfill volume
\therefore Total landfill volume =\frac{65454.5454}{0.8}m^{3}/year =81818.18175m^{3}/year
Height of landfill =3\: m
\therefore Area of lanfill=\frac{81818.18175}{3} =27272.7\: m^{2}/year
Question 5 |
The wastewater form a city, containing a high concentration of biodegradable organics, is beingsteadily discharged into a flowing river at a location S. If the rate of aeration of the river water islower than the rate of degradation of the organics, then the dissolved oxygen of the river water
is lowest at the locations S | |
is lowest at a point upstream of the location S | |
remains constant all along the length of the river | |
is lowest at a point downstream of the location S |
Question 5 Explanation:
As given rate of aeration is less than rate of degradation which decreases with time/distance, minimum DO is observed downstream of point of disposal 'S' where both rate of a reaction and degradation becomes equals.
There are 5 questions to complete.