Treatment of Waste Water


Question 1
In the context of water and wastewater treatments, the correct statements are:
A
particulate matter may shield microorganisms during disinfection
B
ammonia decreases chlorine demand
C
phosphorous stimulates algal and aquatic growth
D
calcium and magnesium increase hardness and total dissolved solids
GATE CE 2023 SET-2   Environmental Engineering
Question 1 Explanation: 
Total dissolved solids comprise of inorganic salts, principally calcium, magnesium, potassium, sodium bicarbonates, chlorides, sulfates and some small amount of organic matter that are dissolved in water.
Hardness is due to presence of multivalent cations. In most cases, it is due to the presence of soluble bicarbonates, chlorides and sulfates of calcium and magnesium.
Too much phosphorous can cause increased growth of algae and large aquatic plants. It can result in decreased levels of dissolved oxygen and the process is called 'Eutrophication'.
Ammonia creates a large chlorine demand, as the ammonia is converted to nitrogen gas and the chlorine becomes chloride during disinfection of water by chlorination.
Presence of particulate matter decreases the effectiveness of UV disinfection or chlorine disinfection by shielding the targeted microorganisms.
Question 2
The wastewater inflow to an activated sludge plant is 0.5 \mathrm{~m}^{3} / \mathrm{s}, and the plant is to be operated with a food to microorganism ratio of 0.2 \mathrm{mg} / \mathrm{mg}-\mathrm{d}. The concentration of influent biodegradable organic matter of the wastewater to the plant (after primary settling) is 150 \mathrm{mg} / \mathrm{L}, and the mixed liquor volatile suspended solids concentration to be maintained in the plant is 2000 \mathrm{mg} / \mathrm{L}. Assuming that complete removal of biodegradable organic matter in the tank, the volume of aeration tank (in \mathrm{~m}^{3} , in integer) required for the plant is _____.
A
16200
B
14752
C
23546
D
54231
GATE CE 2023 SET-1   Environmental Engineering
Question 2 Explanation: 
Given,
\begin{aligned} Q_1&=6.5 m^3/s \\ &=0.5 \times 24 \times 60 \times 60 \\ &=43200 m^3/d\\ \frac{F}{M} & =0.2 (mg/mg-d)\\ \mathrm{S}_0 & =150 \mathrm{mg} / \mathrm{l} \\ \mathrm{X} & =2000 \mathrm{mg} / \mathrm{l} \\ \because \quad \frac{F}{M} & =\frac{\mathrm{Q}_0 \mathrm{~S}_0}{\mathrm{VX}} \\ 0.2 & =\frac{43200 \times 150}{\mathrm{~V} \times 2000} \\ V & =16200 \mathrm{~m}^3\end{aligned}


Question 3
Identify the waterborne diseases caused by viral pathogens:
A
Acute anterior poliomyelitis
B
Cholera
C
Infectious hepatitis
D
Typhoid fever
GATE CE 2023 SET-1   Environmental Engineering
Question 3 Explanation: 
Water Borne Disease caused by

(i) Bacterial -- (a) Typhoid (b) Cholera

(ii) Virus -- (a) Jaundice (Hapatitis virus) (b) Poliomyetitis

(iii) Protozoa -- (a) Amoebic dysentry
Question 4
A 2% sewage sample (in distilled water) was incubated for 3 days at 27^{\circ}C temperature. After incubation, a dissolved oxygen depletion of 10 mg/L was recorded. The biochemical oxygen demand (BOD) rate constant at 27^{\circ}C was found to be 0.23 \; day^{-1} (at base e).
The ultimate BOD (in mg/L) of the sewage will be_____________. (round off to the nearest integer)
A
825
B
1254
C
1003
D
1502
GATE CE 2022 SET-1   Environmental Engineering
Question 4 Explanation: 
Depletion of dissolved oxygen after 3 days incubation = 10 mg/L
\text{Dilution factor }=\frac{\text{Vol. of dilutedsample}}{\text{Vol.of undiluted sewage sample}}=\frac{100}{2}50
BOD_3= \text{DO consumed} \times \text{dilution factor}=10 \times 50=500 mg/L
\begin{aligned} BOD_3&= BOD_u(1-e^{-kt}) \\ 500 &=BOD_u(1-e^{-0.23 \times 3}) \\ BOD_u &= 1003.162 mg/L \end{aligned}
Question 5
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 5 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}


There are 5 questions to complete.

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