Q30GATE 2014MCQ

16 MLD of water is flowing through a 2.5 km long pipe of diameter 45 cm. The chlorine at the rate of 32 kg/d is applied at the entry of this pipe so that disinfected water is obtained at the exit. There is a proposal to increase the flow through this pipe to 22 MLD from 16 MLD. Assume the dilution coefficient, n = 1. The minimum amount of chlorine (in kg per day) to be applied to achieve the same degree of disinfection for the enhanced flow is

🚀 Solve in Practice Mode

📖 Explanation

In the disinfection process we have the relationship,

\begin{aligned} tC^{n} &=K \\ \text{where}\; \;t &=\text{time required to kill all organism} \\ \text{c} &=\text{concentration of disinfectant} \\ n &= \text{dilution coefficient}\\ k &=\text{constant} \\ \Rightarrow \; \; t_{1}C_{1}^{n} &=t_{2}C_{2}^{n} \\ \text{\in our case}\; n &=1 \\ \Rightarrow \; \; t_{1}C_{1}&=t_{2}C_{2} \\ t_{1} &=\frac{L}{v_{1}} \end{aligned}

$L= \text{length of pipe};$ $V_{1}=\text{velocity of flow}$

\begin{aligned} t_{1} &=\frac{L}{Q_{1}/A} \\ t_{1} &=\frac{L.A}{Q_{1}} \\ C_{1} &=\frac{W_{1}}{Q_{1}}, \end{aligned}

\begin{aligned} \text{where } W_{1} &=\text{weight of disinfectant per day}; \\ Q_{1}&=\text{discharge per day} \end{aligned}

\begin{aligned} \Rightarrow \; \; \frac{L.A}{Q_{1}}\times \frac{W_{1}}{Q_{1}} &=\frac{L.A}{Q_{2}}\times \frac{W_{2}}{Q_{2}} \\ \Rightarrow \; \; W_{2} &=\frac{Q_{2}^{2}}{Q_{1}^{2}}\times W_{1} \\ &=\left ( \frac{22}{16} \right )^{2}\times 32\; \text{kg/day} \\ &=60.5\; \text{kg/day} \end{aligned}

Q37GATE 2011MCQ

Chlorine gas (8 mg/L as $CI_2$ ) was added to a drinking water sample. If the free chlorine residual and pH was measured to be 2mg/L (as $CI_2$ ) and 7.5, respectively, what is the concentration of residual $OCI^{-}$ ions in the water? Assume that the chlorine gas added to the water is completely converted to HOCl and $OCI^{-}$ . Atomic Weight of Cl: 35.5 Given: $OC1^{-}+H^{+}\leftrightharpoons^k HOCI, \; \; \; K=10^{7.5}$

🚀 Solve in Practice Mode

📖 Explanation

$OCl^{-}\: +\: H^{+}\: \rightleftharpoons ^{K}\; \; HOCl;\; K=10^{7.5}$ Now,

\begin{aligned} \frac{\left[ HOCl \right]}{\left[ OCl^{-} \right]\$\left[ H^{+} \right]\$} &= K\\ pH&=7.5 \\ \left[ H^{+} \right]\$&=10^{7.5} \\ \left[ \frac{HOCl}{OCl^{-}} \right]\$&=10^{7.5}\times 10^{-7.5}=1 \\ \therefore \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \; \left[ HOCl \right] &=\left[ OCl^{-} \right]\$ \\ \left[ \text{Free chlorine residual} \right]&=\left[ HOCl \right]+\left[ OCl^{-} \right]\$ \\ \text{Free chlorine residual} &= 2\; \text{mg/l}\\ \left[ \text{Free chlorine residual} \right] &=\frac{2\times 10^{-3}}{Mol.\; wt\; of\; Cl_{2}} \\ &=\frac{2\times 10^{-3}}{2\times 35.5} \\ &=2.817\times 10^{-5}\; \text{moles/l} \\ \therefore \; \; \; \; \; \; \; \; 2.817\times 10^{-5} &=\left[ HOCl \right]+\left[ OCl^{-} \right]\$ \\ \therefore \; \; \; \; \; \; \; \; \; \; \left[ OCl^{-} \right]\$ &=\frac{2.817\times 10^{-5}}{2} \\ \because \; \; \; \; \; \; \; \left[ OCl^{-} \right]\$ &=\left[ HOCl \right] \\ &=1.4084\times 10^{-5}\; \text{moles/l} \end{aligned}

Q40GATE 2007MCQ

A plain sedimentation tank with a length of 20 m, width of 10 m, and a depth of 3 m is used in a water treatment plant to treat 4 million litres of water per day (4 MLD). The average temperature of water is 20 $^{\circ}$ C. The dynamic viscosity of water is $1.002 \times 10^{-3}N.s/m^2$ at 20 $^{\circ}$ C. Density of water is 998.2 $kg/m^3$ . Average specific gravity of particles is 2.65. What is the minimum diameter of the particle which can be removed with 100% efficiency in the above sedimentation tank ?

🚀 Solve in Practice Mode

📖 Explanation

Settling velocity,

\begin{aligned} V_{s} &=\frac{H}{t} \\ \text{But} \; \; \; \; \; \; \; t&=\frac{Volume\; of\; tank}{Rate\; of\; flow} \\ &=\frac{20\times 10\times 3\times 24\times 60\times 60}{4\times 10^{6}\times 10^{-3}} \\ &=12960\; \text {seconds} \\ \therefore \; \; \; \; \; \; \; V_{s} &=\frac{3\times 1000}{12960}=0.2315\; \text{mm/s} \end{aligned}

Let the velocity of the particle which can be removed with 100% efficiency be ${V}'_{s}$ then

\begin{aligned} \frac{{V}'_{s}}{V_{s}}\times 100 &=100 \\ \Rightarrow {V}'_{s} &=V_{s} \\ \frac{g}{18}\left ( S_{s}-1 \right )\frac{d^{2}\rho }{\mu } &=0.2315\times 10^{-3} \\ \frac{9.81}{18}\left ( 2.65-1 \right )d^{2}\times \frac{998.2}{1.002\times 10^{-3}} &=0.2315\times 10^{-3} \\ d^{2}&=2.58\times 10^{-10} \\ d &=1.6\times 10^{-5}\text{m} \\ &=16\times 10^{-3}\text{mm} \end{aligned}