Treatment of Waste Water

Question 1
A grit chamber of rectangular cross-section is to be designed to remove particles with diameter of 0.25 mm and specific gravity of 2.70. The terminal settling velocity of the particles is estimated as 2.5 cm/s. The chamber is having a width of 0.50 m and has to carry a peak wastewater flow of 9720 \mathrm{~m}^{3} / \mathrm{d} giving the depth of flow as 0.75 m. If a flow-through velocity of 0.3 m/s has to be maintained using a proportional weir at the outlet end of the chamber, the minimum length of the chamber (in m,in integer) to remove 0.25 mm particles completely is _________
A
5
B
7
C
9
D
10
GATE CE 2021 SET-2   Environmental Engineering
Question 1 Explanation: 
Minimum length of chamber
\begin{aligned} L_{\min } &=V_{\text {flow }} \times t_{d} \\ V_{\text {fow }} &=0.30 \mathrm{~m} / \mathrm{s} \\ t_{d} &=\frac{H}{V_{\text {settling }}}=\frac{0.75}{2.5 \times 10^{-2}}=30 \mathrm{sec} \\ L_{\text {min }} &=0.3 \mathrm{~m} / \mathrm{sec} \times 30 \mathrm{sec} \\ &=9 \mathrm{~m} \end{aligned}
Question 2
An activated sludge process (ASP) is designed for secondary treatment of 7500 m^{3} / day of municipal wastewater. After primary clarifier, the ultimate BOD of the influent, which enters into ASP reactor is 200 mg/L. Treated effluent after secondary clarifier is required to have an ultimate BOD of 20 mg/L. Mix liquor volatile suspended solids (MLVSS) concentration in the reactor and the underflow is maintained as 3000 mg/L and 12000 mg/L, respectively. The hydraulic retention time and mean cell residence time are 0.2 day and 10 days, respectively. A representative flow diagram of the ASP is shown below.

The underflow volume (in m^{3} / day, round off to one decimal place) of sludge wastage is _________
A
25.4
B
37.5
C
65.2
D
87.6
GATE CE 2021 SET-2   Environmental Engineering
Question 2 Explanation: 


\begin{aligned} \text{Given,} \qquad X &=3000 \mathrm{mg} / l, \quad X_{\mathrm{u}}=12000 \mathrm{mg} / l \\ \text{Since} \qquad \mathrm{HRT} &=\frac{V}{Q_{0}}\\ \Rightarrow &\text{Volume of reactor,}\\ V &=Q_{0} \times H R T \\ &=7500 \times 0.2 \mathrm{~m}^{3} \\ &=1500 \mathrm{~m}^{3} \\ &\text{Sludge age,} \\ \theta_{\mathrm{C}} &=\frac{V X}{\left(Q_{0}-Q_{w}\right) X_{e}+Q_{w} X_{u}} \qquad \left(X_{e} \simeq 0\right)\\ 10&=\frac{1500 \times 3000}{Q_{W} \times 12000}\\ Q_{w}&=37.5 \mathrm{~m}^{3} / day \end{aligned}
Question 3
A secondary clarifier handles a total flow of 9600 \mathrm{~m}^{3} / \mathrm{d} from the aeration tank of a conventional activated-sludge treatment system. The concentration of solids in the flow from the aeration tank is 3000 mg/L. The clarifier is required to thicken the solids to 12000 mg/L, and hence it is to be designed for a solid flux of 3.2 \frac{\mathrm{kg}}{\mathrm{m}^{2} \cdot \mathrm{h}}. The surface area of the designed clarifier for thickening (in \mathrm{~m}^{2}, in integer) is _________________
A
125
B
258
C
525
D
375
GATE CE 2021 SET-1   Environmental Engineering
Question 3 Explanation: 
\begin{aligned} Q &=9600 \mathrm{~m}^{3} / \mathrm{d} \\ X &=3000 \mathrm{mg} / l \\ X_{u} &=12000 \mathrm{mg} / l \\ \text {Solid flux }& =3.2 \mathrm{~kg} / \mathrm{m}^{2}-\mathrm{h}\\ \text{Surface area,} \qquad \mathrm{A}&=\frac{\text { Total quantity of solids entering }}{\text { Solid flux }}\\ A &=\frac{9600 \mathrm{~m}^{3} / \mathrm{d} \times 3000 \mathrm{mg} / l}{3.2 \times 10^{6} \mathrm{mg} / \mathrm{m}^{2}-\mathrm{h}} \\ &=\frac{9600 \times 10^{3} \mathrm{lld} \times 3000 \mathrm{mg} / \mathrm{l}}{3.2 \times 10^{6} \mathrm{mg} / \mathrm{m}^{2}-\mathrm{h} \times 24 \mathrm{~h} / \mathrm{d}} \\ &=375 \mathrm{~m}^{2} \end{aligned}
Question 4
A wastewater is to be disinfected with 35 mg/Lof chlorine to obtain 99% kill of micro-organisms. The number of micro-organisms remaining alive (N_t) at time t, is modelled by N_t=N_oe^{-kt}, where N_o is number of micro-organisms at t= 0, and k is the rate of kill. The wastewater flow rate is 36 m^3/hr, and k= 0.23 min^{-1}. If the depth and width of the chlorination tank are 1.5 m and 1.0 m, respectively, the length of the tank (in m, round off to 2 decimal places) is ________
A
2.65
B
8.65
C
8.00
D
6.74
GATE CE 2019 SET-1   Environmental Engineering
Question 4 Explanation: 
\begin{aligned} N_t&=N_0e^{-kt}\\ \therefore \; \eta &=[1-e^{-kt}]100\\ 99&=[1-e^{-0.23\times t}]100\\ \therefore \;t&=20.22min\\ Q&=36m^3/hr\\ V&=Q \times t\\ &=\frac{36}{60}\times 20.22=12.0132\\ \text{Length}&=\frac{V}{B \times H}=\frac{12.0132}{1 \times 1.5}\\ &=8.008\cong 8m \end{aligned}
Question 5
A schematic flow diagram of a completely mixed biological reactor with provision for recycling of solids is shown in the figure.

The mean cell residence time (in days, up to one decimal place) is ______
A
2.5
B
5.5
C
7.5
D
10.2
GATE CE 2018 SET-2   Environmental Engineering
Question 5 Explanation: 
\begin{aligned} \theta_{C} &=\frac{V X}{\left(Q_{0}-Q_{w}\right) X e+Q_{w} X_{u}} \\ &=\frac{Q_{0} H R T X}{\left(Q_{0}-Q_{w}\right) X_{e}+Q_{w} X_{u}}\left(X_{e}=0\right)\\ &=\frac{15000 \times \frac{2}{24} \times 3000}{50 \times 10000}=7.5 \text{ days} \end{aligned}\\
Question 6
At a small water treatment plant which has 4 filters, the rates of filtration and backwashing are 200 m^{3}/d/m^{2} and 1000 m^{3}/d/m^{2}, respectively. Backwashing is done for 15 min per day. The maturation, which occurs initially as the filter is put back into service after cleaning, takes 30 min. It is proposed to recover the water being wasted during backwashing and maturation. The percentage increase in the filtered water produced (up to two decimal places) would be ______
A
7.52
B
2.25
C
8.22
D
9.85
GATE CE 2018 SET-2   Environmental Engineering
Question 6 Explanation: 
Let total area of filters be 1 \mathrm{m}^{2}
Water used for backwashing
=1000 \times \frac{15}{24 \times 60}=10.4166 \mathrm{m}^{3}
Water used for maturation
=200 \times \frac{30}{24 \times 60}=4.166 \mathrm{m}^{3}
Total water wasted for backwashing and
maturation
=10.4166+4.166=14.58 \mathrm{m}^{3}
Water to be treated by filtered
=200 \times \frac{23.25}{24}=193.75 \mathrm{m}^{3} / \mathrm{day}
% increase in filtered water produced
=\frac{14.58}{193.75} \times 100=7.525 \%
Question 7
A waste activated sludge (WAS) is to be blended with green waste (GW). The carbon (C) and nitrogen (N) contents, per kg of WAS and GW, on dry basis are given in the table.

The ratio of WAS to GW required (up to two decimal places) to achieve a blended C:N ratio of 20:1 on dry basis is ______
A
1.26
B
1.94
C
2.65
D
1.64
GATE CE 2018 SET-1   Environmental Engineering
Question 7 Explanation: 
Let 20 kg of C and 1 kg of N is required
Let x kg of WAS is taken
\begin{aligned} \therefore \quad \text{Carbon in x } \mathrm{kg}&=0.054 x \mathrm{kg} \\ \text{ Nitrogen in x } \mathrm{kg}&=0.010 x \mathrm{kg} \end{aligned}
Let y kg of GW is taken
\begin{aligned} \therefore \quad \text { Carbon in } y \mathrm{kg} &=0.360 \mathrm{y} \mathrm{kg} \\ \text { Nitrogen in } y \mathrm{kg} &=0.006 \mathrm{y} \mathrm{kg} \end{aligned}
Total Carbon,
0.054 x+0.36 y=20 \mathrm{kg}\quad \quad\ldots(i)
Total Nitrogen,
\begin{aligned} 0.01 x+0.006 y &=1 \mathrm{kg} &\ldots(ii)\\ x &=73.26 \mathrm{kg} \\ y &=44.566 \mathrm{kg} \\ \frac{x}{y} &=\frac{\mathrm{WAS}}{\mathrm{GW}}\\ \therefore &=1.6438 \simeq 1.64 \end{aligned}
Question 8
The most important type of species involved in the degradation of organic matter in the case of activated sludge process is
A
Autotrophs
B
Heterotrophs
C
Prototrophs
D
Photo-autotrophs
GATE CE 2017 SET-2   Environmental Engineering
Question 8 Explanation: 
Activated sludge process is designed primarily for satisfaction of carbonaceous BOD which is done by heterotrophs.
Question 9
The wastewater having an organic concentration of 54 mg/l is flowing at a steady rate of 0.8 m^{3}/day through a detention tank of dimensions 2m\times 4m\times 2m. If the contents of the tankare well mixed and the decay constant is 0.1 per day, the outlet concentration (in mg/l, up to one decimal place) is ______
A
17
B
21
C
20
D
18
GATE CE 2017 SET-1   Environmental Engineering
Question 9 Explanation: 
Q=0.8 \mathrm{m}^{3} / \mathrm{d}
Detention time,
\begin{aligned} D_{t}&=\frac{V}{Q}=\frac{4 \times 2 \times 2 \mathrm{m}^{3}}{0.8 \mathrm{m}^{3} / \mathrm{day}}=20 \mathrm{days} \\ L_{t}&=\frac{L_{0}}{1+k_{\mathrm{e}} t}=\frac{54}{1+0.1 \times 20}=18 \mathrm{mg} / l \end{aligned}
Question 10
Match the items in Group-I with those in Group-II and choose the right combination.
A
P-3, Q-4, R-2, S-1
B
P-2, Q- R-4, S-1
C
P-3, Q-2, R-4, S-1
D
P-1, Q-4, R-2, S-3
GATE CE 2016 SET-1   Environmental Engineering
There are 10 questions to complete.

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