Water Treatment Practice Questions

We are given: TDS = 500 mg/L TH = 400 mg/L (as CaCO3) Volume = 1 L Only two salts: $\text{MgCl}_2$ and $\text{CaSO}_4$ Molecular weight of $\text{MgCl}_2$ = $24 + (35.5 \times 2) = 95\ \text{g/mol}$ Molecular weight of $\text{CaSO}_4$ = $40 + 32 + (16 \times 4) = 136\ \text{g/mol}$ Equivalent weight of $\text{Mg}^{2+}$ = $\frac{24}{2} = 12$ Equivalent weight of $\text{Ca}^{2+}$ = $\frac{40}{2} = 20$ Let x = mg of $\text{MgCl}_2$ Let y = mg of $\text{CaSO}_4$ TDS equation: $x + y = 500$ TH from $\text{MgCl}_2$ = $\left( \frac{24}{95}x \right) \times \frac{50}{12} = \frac{100x}{95}$ TH from $\text{CaSO}_4$ = $\left( \frac{40}{136}y \right) \times \frac{50}{20} = \frac{100y}{136}$ Total TH = $\frac{100x}{95} + \frac{100y}{136} = 400$ Divide by 100: $\frac{x}{0.95} + \frac{y}{1.36} = 4 \quad \text{(1)}$ $x + y = 500 \quad \text{(2)}$ From (2): $y = 500 - x$ Substitute into (1): $\frac{x}{0.95} + \frac{500 - x}{1.36} = 4$ Try x = 103, y = 397 Check TDS: 103 + 397 = 500 Mg hardness: $\frac{24}{95} \times 103 \times \frac{50}{12} \approx 108.7$ Ca hardness: $\frac{40}{136} \times 397 \times \frac{50}{20} \approx 291.3$ Total hardness = 400 Final Answer: (C) Amount of $\text{MgCl}_2$ added is 103 mg (D) Amount of $\text{CaSO}_4$ added is 397 mg

$\text{Initial FC concentration} = 10780 \ \text{cfu/100 ml}$ $\text{Final FC concentration (after treatment)} = 23 \ \text{cfu/100 ml}$ $\text{Log Kill} = \log_{10} \left( \frac{\text{Initial concentration}}{\text{Final concentration}} \right)$ $\quad = \log_{10} \left( \frac{10780}{23} \right)$ $\quad = \log_{10} (468.7)$ $\quad \approx 2.67$

Given: Surface Overflow Rate (SOR) = $45 \, \text{m}^3/\text{m}^2/\text{day} = \frac{45}{86400} \, \text{m/s} \approx 5.208 \times 10^{-4} \, \text{m/s}$ Specific gravity of inorganic particles $(G_{\in}) = 2.65$ Specific gravity of organic particles $(G_{org}) = 1.20$ Acceleration due to gravity $g = 9.81 \, \text{m/s}^2$ Kinematic viscosity $\nu = 1 \times 10^{-6} \, \text{m}^2/\text{s}$ Using Stoke?s Law for settling velocity: $V_s = \frac{g (G - 1) d^2}{18 \nu}$ Setting $V_s = \text{SOR} = 5.208 \times 10^{-4} \, \text{m/s}$ Solving for $d$ : $d = \sqrt{ \frac{18 \nu V_s}{g (G - 1)} }$ For inorganic particles (G = 2.65): $d = \sqrt{ \frac{18 \times 1 \times 10^{-6} \times 5.208 \times 10^{-4}}{9.81 \times (2.65 - 1)} }$ $d \approx \sqrt{ \frac{9.3744 \times 10^{-9}}{16.0665} } \approx \sqrt{5.83 \times 10^{-10}} \approx 2.414 \times 10^{-5} \, \text{m} = 24.14 \, \mu\text{m}$ For organic particles (G = 1.20): $d = \sqrt{ \frac{18 \times 1 \times 10^{-6} \times 5.208 \times 10^{-4}}{9.81 \times (1.20 - 1)} }$ $d \approx \sqrt{ \frac{9.3744 \times 10^{-9}}{1.962} } \approx \sqrt{4.78 \times 10^{-9}} \approx 6.91 \times 10^{-5} \, \text{m} = 69.1 \, \mu\text{m}$ Correct options: (A) Minimum diameter of inorganic particles is 24 ?m (B) Minimum diameter of organic particles is 69 ?m

Given: Chlorine concentration = 2 mg/l (as $Cl_2$ ) pH = 8.5 pK = 7.5 Molecular weight of $Cl_2 = 71$ g/mol Step 1: Convert to molar concentration $\text{Moles of } Cl_2 = \frac{2}{71 \times 10^3} = 2.82 \times 10^{-5} \ \text{mol/L}$ Each mole of $Cl_2$ gives 2 moles of total chlorine species ( $HOCl$ and $OCl^-$ ): [HOCl] + \[OCl^-]$ = 2 \times 2.82 \times 10^{-5} = 5.63 \times 10^{-5} \ \text{mol/L}$Step 2: Use Henderson-Hasselbalch equation$\text{pH} = \text{p}K + \log_{10} \left( \frac{[OCl^-]$}{[HOCl]} \right)$$8.5 = 7.5 + \log_{10} \left( \frac{[OCl^-]$}{[HOCl]} \right)$$\frac{[OCl^-]$}{[HOCl]} = 10$Let$[HOCl] = x$, then$[OCl^-] = 10x$$x + 10x = 5.63 × 10^{-5} \Rightarrow 11x = 5.63 × 10^{-5}$$x = \frac{5.63 × 10^{-5}}{11} = 5.12 × 10^{-6} \ \text{mol/L}$Step 3: Convert to umol/L$[HOCl] = 5.12 × 10^{-6} \ \text{mol/L} = 5.12 \ \text{umol/L}$But since the concentration was given as$Cl_2$, we adjust for only 1 mole of chlorine contributing to HOCl:$[HOCl] = \frac{2}{71} × 10^6 × \frac{1}{11} = 2.56 \ \text{umol/L}$Final Answer:$\boxed{2.6 \ \text{umol/L}}$

Natural alkalinity $=4 \mathrm{mg} / l$ as $\mathrm{CaCO}_3$

$\Rightarrow$ Alkalinity needed to be provided through outside source $=(11.11-4)=7.11 \mathrm{mg} / l$ as $\mathrm{CaCO}_3$ $=\frac{7.11}{50}$ milieq $/ l$ $\Rightarrow \mathrm{CaO}$ required $=\frac{7.11}{50}$ milieq $/ l$ $=\left[\frac{7.11}{50} \times \right.$ eq. wt of $\left.\mathrm{CaO}\right] \frac{\mathrm{mg}}{l}$ $=\left[\frac{7.11}{50} \times 28\right] \frac{\mathrm{mg}}{l} \times 25 \times 10^6 l / \mathrm{d}$ $=99.54 \mathrm{~kg} / \mathrm{d}$

Let the settling velocity of anthracite pacticle at top, silica pacticle at middle & ilmenite pacticle at bottom, after back washing be $\left(V_{S}\right)_{T},\left(V_{S}\right),\left(V_{S}\right)_{B}$ respectively. For Middle silica sand

$V_{S}=K(1.6)(0.2)^{2} \text { where, } K=\frac{\gamma_{w}}{18 \mu}$ For Top Anthracite

As anthracite lies above silica layer So, anthracite particles should have settling velocity less than silica particles $\left(\mathrm{V}_{\mathrm{S}}\right)_{\mathrm{T}} \lt \mathrm{V}_{\mathrm{S}}$ So, $K(0.5)\left(D_{2}\right)^{2} \lt K(1.6)(0.2)^{2}$ $\mathrm{D}_{2}$ or $\mathrm{D}_{\text {Top }} \lt \sqrt{\frac{1.6 \times (0.2)^{2}}{0.5}}$ $\mathrm{D}_{2}$ or $\mathrm{D}_{\text {Top }} \lt 0.357 \mathrm{~mm}$ For bottom ilmenite sand $\left(\mathrm{V}_{\mathrm{S}}\right)_{\mathrm{B}}=\mathrm{K}\left(\mathrm{G}_{\mathrm{S}_{3}}-1\right)\left(\mathrm{D}_{3}^{2}\right)$ $\left(\mathrm{V}_{\mathrm{S}}\right)_{\mathrm{B}}=\mathrm{K}(4.2-1)\left(\mathrm{D}_{3}^{2}\right)$ As, ilimite layer lies below silica lyers So, ilimite particles should have settling velocity greater than silica i.e., $\left(V_{S}\right)_{B} \gt V_{S}$ $\mathrm{K}(3.2)\left(\mathrm{D}_{3}^{2}\right) \gt \mathrm{K}(1.6)(0.2)^{2}$ $\mathrm{D}_{3}$ or $\mathrm{D}_{\text {bottom }} \gt \sqrt{\frac{1.6 \times (0.2)^{2}}{3.2}}$ $\mathrm{D}_{3}$ or $\mathrm{D}_{\text {bottom }} \gt 0.141 \mathrm{~mm}$ Hence, diameter of anthracite particle is slightly greater than $0.35 \mathrm{~mm} \&$ diameter of ilmenite particle is slightly greater than $0.141 \mathrm{~mm}$

$\mathrm{O}_{3}, \mathrm{OH}, \mathrm{OCl}^{-}$ 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 $\left.\mathrm{HOCl}\right)$ .

Given, Volume of flocculation tank $(V)=2800 \mathrm{~m}^{3}$ At $\mathrm{T}=15^{\circ} \mathrm{C}$

At $\mathrm{T}=5^{\circ} \mathrm{C}$

$\because$ 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

Currently no explanation available

Desalination can be done by Electrodialysis Reverse osmosis

Hardness end point of titration --> wine red to blu

Waterson law,

$Cl_2+ H_2O\overset{pH \gt 5}{\rightleftharpoons } HOCl+HCl$ $HOCl \xrightleftharpoons [pH \lt 7]{pH \gt 8} H^+ +OCl^-$ At $pH \lt 5$ , chlorine does not react with water and remains as free chlorine. $(HOCl^- +OCl^- +cl_2)$ are combinedly called freely available chlorine. Out of these forms of freely available chlorine, $HOCl$ is most destructive. It is 80% more effective than $OCl^-$ ion. Hence, pH of water should be maintained slightly below 7.

Given,