Water Treatment


Question 1
Which of the following is/are NOT active disinfectant(s) in water treatment?
A
\cdot \mathrm{OH} (hydroxyl radical)
B
\mathrm{O}_{3} (ozone)
C
\mathrm{OCl}^{-}(hypochlorite ion)
D
\mathrm{Cl}^{-}(chloride ion)
GATE CE 2023 SET-2      Water Treatment
Question 1 Explanation: 
\mathrm{O}_{3}, \mathrm{OH}, \mathrm{OCl}^{-}are considered to be the active disinfectants in water treatment.
NOTE:
1. \left(\mathrm{OH}^{-}\right)Hydroxyl ion is not an active disinfectant.
2. \left(\mathrm{Cl}^{-}\right)chloride ion reduces disinfection efficiency of free chlorine \left(\mathrm{OCl}^{-}\right.and \left.\mathrm{HOCl}\right).
Question 2
A flocculator tank has a volume of 2800 \mathrm{~m}^{3}. The temperature of water in the tank is 15^{\circ} \mathrm{C}, and the average velocity gradient maintained in the tank is 100 / \mathrm{s}. The temperature of water is reduced to 5^{\circ} \mathrm{C}, but all other operating conditions including the power input are maintained as the same. The decrease in the average velocity gradient (in %) due to the reduction in water temperature is ____ (round off to nearest integer).
[Consider dynamic viscosity of water at 15^{\circ} \mathrm{C} and 5^{\circ} \mathrm{C} as 1.139 \times 10^{-3} \mathrm{~N}-\mathrm{s} / \mathrm{m}^{2} and 1.518 \times 10^{-3} \mathrm{~N}-\mathrm{s} / \mathrm{m}^{2}, respectively]
A
10
B
13
C
17
D
22
GATE CE 2023 SET-1      Water Treatment
Question 2 Explanation: 
Given,
Volume of flocculation tank (V)=2800 \mathrm{~m}^{3}
At \mathrm{T}=15^{\circ} \mathrm{C}
\begin{aligned} & \mathrm{G}_{15^{\circ} \mathrm{C}}=100 / \mathrm{s}, \\ & \eta_{15^{\circ} \mathrm{C}}=1.139 \times 10^{-3} \mathrm{~N}-\mathrm{s} / \mathrm{m}^{2} \end{aligned}
At \mathrm{T}=5^{\circ} \mathrm{C}
\begin{aligned} & \mathrm{G}_{5^{\circ} \mathrm{C}}=? \\ & \mu_{5^{\circ} \mathrm{C}}=1.518 \times 10^{-3} \mathrm{~N}-\mathrm{s} / \mathrm{m}^{2} \end{aligned}
\because Temporal mean velocity gradient,
G=\sqrt{\frac{P}{\mu V}}
\mathrm{P} and \mathrm{V} are constant.
So, \quad G^{2} \mu= constant
100^{2} \times 1.139 \times 10^{-3}=G_{5^{\circ} \mathrm{C}}^{2} \times 1.518 \times 10^{-3}
\mathrm{G}_{5^{\circ} \mathrm{C}}=86.62 \mathrm{sec}^{-1}
The decrease in velocity gradient
\begin{aligned} & =\frac{G_{15^{\circ} \mathrm{C}}-\mathrm{G}_{5^{\circ} \mathrm{C}}}{\mathrm{G}_{15^{\circ} \mathrm{C}}} \times 100 \\ & =13.38 \% \end{aligned}


Question 3
In the following table, identify the correct set of associations between the entries in Column-1 and Column-2.
\begin{array}{|l|l|} \hline \text{Column-1} & \text{Column-2} \\ \hline \text{P : Reverse Osmosis} & \text{I: Ponding} \\ \hline \text{Q: Trickling Filter} & \text{II : Freundlich Isothrem} \\ \hline \text{R : Coagulation} &\text{ III : Concentration Polarization} \\ \hline \text{S : Adsorption} & \text{IV : Charge Neutralization} \\ \hline \end{array}
A
P-II, Q-I, S-III
B
Q-III, R-II, S-IV
C
P-IV, R-I, S-II
D
P-III, Q-I, R-IV
GATE CE 2023 SET-1      Water Treatment
Question 4
A water treatment plant has a sedimentation basin of depth 3 m, width 5 m, and length 40 m. The water inflow rate is 500 m^3/h . The removal fraction of particles having a settling velocity of 1.0 m/h is_______. (round off to one decimal place)
(Consider the particle density as 2650 kg/m^3 and liquid density as 991 kg/m^3 )
A
0.4
B
0.2
C
0.1
D
0.8
GATE CE 2022 SET-1      Water Treatment
Question 4 Explanation: 
\begin{aligned} V_o&=\frac{500}{5 \times 40} =2.5m/h \\ \eta &=\frac{V_s}{V_o}\times 100=\frac{1}{2.5}=0.4 \; or \; 40% \end{aligned}
Question 5
Which of the following process(es) can be used for conversion of salt water into fresh water?
A
Microfiltration
B
Electrodialysis
C
Ultrafiltration
D
Reverse osmosis
GATE CE 2022 SET-1      Water Treatment
Question 5 Explanation: 
Desalination can be done by
Electrodialysis
Reverse osmosis


There are 5 questions to complete.

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