GATE CE · Environmental Engineering

Water Treatment Practice Questions

Generate GATE-level questions on Water Treatment in Environmental Engineering. Focus on core concepts, previous year patterns, and numerical problem-solving techniques.

52 questions · 12 PYQs · 0 AI practice · GATE CE 2027

Q41GATE 2006MCQ

In a rapid sand filter, the time for reaching particle break through ( $T_B$ ) is defined as the time elapsed from start of filter run to the time at which the turbidity of the effluent from the filter is greater than 2.5 NTU. The time for reaching terminal head loss ( $T_H$ ) is defined as the time elapsed from the start of the filter run to the time when head loss across the filter is greater than 3 m. The effect of increasing the filter loading rate (while keeping all other conditions same) on $T_{B}$ and $T_H$ is

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Q42GATE 2005MCQ

A circular primary clarifier processes an average flow of 5005 $m^{3}$ /d of municipal waste water. The overflow rate is 35 $m^{3}/m^{2}$ /d. The diameter of clarifier shall be

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📖 Explanation

\begin{aligned} \text{Area of clarifier} &=\frac{\text{Average flow of waste water}}{\text{Overflow rate}} \\ &=\frac{5005}{35}=143\; m^{2} \\ \therefore \; \; \frac{\pi }{4}D^{2} &=143 \\ \Rightarrow \; \; D^{2} &=\frac{143\times 4}{\pi } \\ \Rightarrow \; \; D &=13.5\; \text{m} \end{aligned}

Q46GATE 2003MCQ

An ideal horizontal flow settling basing is 3 m deep having surface area 900 $m^{2}$ . Water flows at the rate of 8000 $m^{3}/d$ , at water temperature 20 $^{\circ}$ C (m= $10^{-3}$ kg/m.s and $\rho =1000 kg/m^3$ ). Assuming Stoke's law to be valid, the proportion (percentage) of spherical sand particles (0.01 mm in diameter with specific gravity 2.65), that will be removed, is

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📖 Explanation

Velocity of sand particles,

\begin{aligned} {V}'_{s} &=\frac{g}{18}\left ( S_{s}-1 \right )\frac{d^{2}}{\nu } \\ &=\frac{9.81}{18}\times \left ( 2.65-1 \right )\times \frac{\left ( 0.001\times 10^{-3} \right )^{2}}{\left ( 10^{-3}/1000 \right )} \\ &=\frac{9.81}{18}\times 1.65\times 10^{-4} \\ {V}'_{s} &=8.9925\times 10^{-5}\; \text{m/sec} \end{aligned}

$\text{settling time}$

\begin{aligned} &=\frac{\text{Volume of tank}}{\text{Water flow rate}} \\ &=\frac{900\times 3}{8000}days=\frac{2700}{8000}\times 24=8.1\; hrs \end{aligned}

$\text{Settling velocity,}$

\begin{aligned} V_{s} &=\frac{\text{Depth of tank}}{\text{Settling time}}=\frac{3}{8.1\times 3600} \\ &=1.0288\times 10^{-4}\; \text{m/s} \end{aligned}

$\therefore \; \;\text{ Removal efficiency}=\frac{{V}'_{s}}{V_{s}}\times 100$

\begin{aligned} &=\frac{8.9925\times 10^{-5}}{1.0288\times 10^{-4}}\times 100 \\ &=8.41\%\simeq 87.5\% \end{aligned}