Question 1 |
Which of the following is/are NOT active disinfectant(s) in water treatment?
\cdot \mathrm{OH} (hydroxyl radical) | |
\mathrm{O}_{3} (ozone) | |
\mathrm{OCl}^{-}(hypochlorite ion) | |
\mathrm{Cl}^{-}(chloride ion) |
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).
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]
[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]
10 | |
13 | |
17 | |
22 |
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}
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}
\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}
P-II, Q-I, S-III | |
Q-III, R-II, S-IV
| |
P-IV, R-I, S-II | |
P-III, Q-I, R-IV |
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 )
(Consider the particle density as 2650 kg/m^3 and liquid density as 991 kg/m^3 )
0.4 | |
0.2 | |
0.1 | |
0.8 |
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?
Microfiltration | |
Electrodialysis | |
Ultrafiltration | |
Reverse osmosis |
Question 5 Explanation:
Desalination can be done by
Electrodialysis
Reverse osmosis
Electrodialysis
Reverse osmosis
There are 5 questions to complete.