Question 1 |
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 1 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 2 |
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 2 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 3 |
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 3 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.
Question 4 |
In a certain situation, waste water discharged into a river mixer with the river
water instantaneously and completely. Following is the data available :
Waste water:
DO = 2.00 mg/L
Discharge rate = 1.10 m^{3}/s
River water:
DO = 8.3 mg/L
Flow rate = 8.70 m^{3}/s
Temperature = 20^{\circ}C
Initial amount of DO in the mixture of waste and river shall be
Waste water:
DO = 2.00 mg/L
Discharge rate = 1.10 m^{3}/s
River water:
DO = 8.3 mg/L
Flow rate = 8.70 m^{3}/s
Temperature = 20^{\circ}C
Initial amount of DO in the mixture of waste and river shall be
5.3 mg/L | |
6.5 mg/L | |
7.6 mg/L | |
8.4 mg/L |
Question 4 Explanation:
\begin{aligned} DO_{mix} &=\frac{Q_{W}DO_{W}+Q_{R}DO_{R}}{Q_{W}+Q_{R}} \\ &=\frac{\left ( 1.10\times 2.00 \right )+\left ( 8.70\times 8.3 \right )}{1.10+8.70} \\ &=7.6\: mg/L \end{aligned}
Question 5 |
Match Group-I (Characteristics of sewage discharged into inland waters ) with Group-II (Allowable limit, mg/L).
P-2 Q-5 R-4 S-2 | |
P-4 Q-1 R-6 S-4 | |
P-3 Q-1 R-4 S-2 | |
P-2 Q-1 R-6 S-3 |
Question 5 Explanation:
There are 5 questions to complete.