# 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
$100^{2} \times 1.139 \times 10^{-3}=G_{5^{\circ} \mathrm{C}}^{2} \times 1.518 \times 10^{-3}$
$\mathrm{G}_{5^{\circ} \mathrm{C}}=86.62 \mathrm{sec}^{-1}$
\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.