Question 1 |

The hardness of a water sample is measured directly by titration with 0.01 M solution of ethylenediamine tetraacetic acid (EDTA) using eriochrome black T (EBT) as an indicator. The EBT reacts and forms complexes with divalent metallic cations present in the water. During titration, the EDTA replaces the EBT in the complex. When the replacement of EBT is complete at the end point of the titration, the colour of the solution changes from

blue-green to reddish brown | |

blue to colourless | |

reddish brown to pinkish yellow | |

Wine red to blue |

Question 1 Explanation:

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

Question 2 |

Surface Overflow Rate (SOR) of a primary settling tank (discrete settling) is 20000 litre/m^2
per day. Kinematic viscosity of water in the tank is 1.01 times 10^{-2} cm^2/s. Specific gravity
of the settling particles is 2.64. Acceleration due to gravity is 9.81 m/s^2. The minimum
diameter (in ?m, round off to one decimal place) of the particles that will be removed
with 80% efficiency in the tank, is _________.

12.14 | |

18.36 | |

22.78 | |

14.46 |

Question 2 Explanation:

\begin{aligned} \eta &=80=\frac{u_s}{v_s}\times 100\\ u_s&=\frac{0.8 \times 20000 \times 10^{-3}}{86400}\\ &=1.85 \times 10^{-4} m/sec\\ u_s&=\frac{(G-1)gd^2}{18v}\\ 1.85 \times 10^{-4}&=\frac{(2.64-1)\times 9.81 \times d^2}{18 \times 1.01 \times 10^{-2} \times 10^{-4}}\\ d&=1.446 \times 10^{-5}m\\ &=14.46\mu m \end{aligned}

Question 3 |

A water supply scheme transports 10 MLD (Million Litres per Day) water through a 450
mm diameter pipeline for a distance of 2.5 km. A chlorine dose of 3.50 mg/litre is applied
at the starting point of the pipeline to attain a certain level of disinfection at the downward
end. It is decided to increase the flow rate from 10 MLD to 13 MLD in the pipeline. Assume
exponent for concentration, n = 0.86. With this increased flow, in order to attain the same
level of disinfection, the chlorine does (in mg/litre) to be applied at the starting point
should be

3.95 | |

4.4 | |

4.75 | |

5.55 |

Question 3 Explanation:

Waterson law,

\begin{aligned} tc^h &=\text{Constant} \\ t_1c_1^n&= t_2c_2^n\\ \frac{d_1}{v_1}c_1^n&=\frac{d_2}{v_2}c_2^n \\ \frac{d_1}{Q_1}A_1c_1^n&=\frac{d_2A_2c_2^n}{Q_2} \\ d_1=d_2, \;& A_1=A_2 \\ \frac{(3.5)^{0.86}}{10} &=\frac{(c_2)^{0.86}}{13} \\ c_2 &=\left ( \frac{13}{10} \right )^{1/0.86} \times 3.5 \\ &= 4.747 mg/l \end{aligned}

\begin{aligned} tc^h &=\text{Constant} \\ t_1c_1^n&= t_2c_2^n\\ \frac{d_1}{v_1}c_1^n&=\frac{d_2}{v_2}c_2^n \\ \frac{d_1}{Q_1}A_1c_1^n&=\frac{d_2A_2c_2^n}{Q_2} \\ d_1=d_2, \;& A_1=A_2 \\ \frac{(3.5)^{0.86}}{10} &=\frac{(c_2)^{0.86}}{13} \\ c_2 &=\left ( \frac{13}{10} \right )^{1/0.86} \times 3.5 \\ &= 4.747 mg/l \end{aligned}

Question 4 |

During chlorination process, aqueous (aq) chlorine reacts rapidly with water to from CI^-, HOCI, and H^+ as shown below

CI_2(aq)+ H_2O\rightleftharpoons HOCI+CI^-+H^+

The most active disinfectant in the chlorination process from amongst the following, is

CI_2(aq)+ H_2O\rightleftharpoons HOCI+CI^-+H^+

The most active disinfectant in the chlorination process from amongst the following, is

H^+ | |

HOCI | |

CI^- | |

H_2O |

Question 4 Explanation:

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.

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.

Question 5 |

A water treatment plant treats 6000 m^3 of water per day. As a part of the treatment process, discrete particles are required to be settled in a clarifier. A column test indicates that an overflow rate of 1.5 m per hour would produce the desired removal of particles through settling in the clarifier having a depth of 3.0 m. The volume of the required clarifier, (in m^3, round off to 1 decimal place) would be _______

225 | |

500 | |

625 | |

725 |

Question 5 Explanation:

\begin{aligned} \text{Design discharge, }\\ Q_0&=6000m^3/d\\ \text{Overflow rate}\\ OFR(v_s)&=1.5m/hr\\ SA&=\frac{Q_0}{OFR}\\ \text{Volume,}\; V&=SA \times \text{Depth}(H)\\ &=\frac{Q_0}{OFR} \times H\\ &=\frac{6000 \times 3}{1.5 \times 24}=500m^3 \end{aligned}

Question 6 |

Chlorine is used as the disinfectant in a municipal water treatment plant. It achieves 50 percent of disinfection efficiency measured in terms of killing the indicator microorganisms (E-Coli) in 3 minutes. The minimum time required to achieve 99 percent disinfection efficiency would be

9.93 minutes | |

11.93 minutes | |

19.93 minutes | |

21.93 minutes |

Question 6 Explanation:

\begin{aligned} \eta _1&=50\%,\; t_1=3 \;min \\ \eta _1 &=[1-e^{-k \times t_1}] \times 100 \\ 50&=[1-e^{-k \times 3}] \times 100\\ k &=0.231 \; min^{-1} \\ \eta _2&=99\%,\; t_2=? \\ \eta _1&= [1-e^{-k \times t_2}] \times 100 \\ 99&= [1-e^{-0.231 \times t_2}] \times 100\\ t_2 &=19.93\; min \end{aligned}

Question 7 |

A 0.80 m deep bed of sand filter (length 4 m and width 3 m) is made of uniform particles (diameter = 0.40 mm, specific gravity = 2.65, shape factor = 0.85) with bed porosity of 0.4. The bed has to be backwashed at a flow rate of 3.60 m^3/min. During backwashing, if the terminal settling velocity of sand particles is 0.05 m/s, the expanded bed depth (in m, round off to 2 decimal places) is ________

1.20 | |

1.75 | |

1.85 | |

0.65 |

Question 7 Explanation:

Given,

\begin{aligned} z &=0.8m \\ n &=0.4m \\ Q_B &=3.6m^3/min \\ &= \frac{3.6}{60}m^3/sec=0.06m^3/sec\\ L&=4m \\ B&= 3m\\ V_B&= \frac{Q_B}{A_s}=\frac{Q_B}{LB}\\ &= \frac{0.06}{4 \times 3}=5 \times 10^{-3}m/sec\\ V_s&= 0.05m/sec\\ V_B&=V_s(n_e)^{4.5} \\ (n_e)^{4.5}&= \frac{V_B}{V_s}=\frac{5 \times 10^{-3}}{0.05}\\ n_e&=\left ( \frac{5 \times 10^{-3}}{0.05} \right )^{\frac{1}{4.5}} \\ &= 0.599\simeq 0.6\\ \frac{z_e}{z}&=\frac{1-n}{1-n_e} \\ z_e &=z\left ( \frac{1-n}{1-n_e} \right ) \\ z_e&=0.8 \times \left ( \frac{1-0.4}{1-0.6} \right )=1.2m \end{aligned}

\begin{aligned} z &=0.8m \\ n &=0.4m \\ Q_B &=3.6m^3/min \\ &= \frac{3.6}{60}m^3/sec=0.06m^3/sec\\ L&=4m \\ B&= 3m\\ V_B&= \frac{Q_B}{A_s}=\frac{Q_B}{LB}\\ &= \frac{0.06}{4 \times 3}=5 \times 10^{-3}m/sec\\ V_s&= 0.05m/sec\\ V_B&=V_s(n_e)^{4.5} \\ (n_e)^{4.5}&= \frac{V_B}{V_s}=\frac{5 \times 10^{-3}}{0.05}\\ n_e&=\left ( \frac{5 \times 10^{-3}}{0.05} \right )^{\frac{1}{4.5}} \\ &= 0.599\simeq 0.6\\ \frac{z_e}{z}&=\frac{1-n}{1-n_e} \\ z_e &=z\left ( \frac{1-n}{1-n_e} \right ) \\ z_e&=0.8 \times \left ( \frac{1-0.4}{1-0.6} \right )=1.2m \end{aligned}

Question 8 |

Sedimentation basin in a water treatment plant is designed for a flow rate of 0.2 m^3/s. The basin is rectangular with a length of 32m, width of 8m, and depth of 4m. Assume that the settling velocity of these particles is governed by the Stokes' law. Given:

density of the particles = 2.5 g/cm^3;

density of water = 1 g/cm^3;

dynamic viscosity of water = 0.01 g/(cm.s);

gravitational acceleration = 980 cm/s^2.

If the incoming water contains particles of diameter 25 \mu m(spherical and uniform), the removal efficiency of these particles is

density of the particles = 2.5 g/cm^3;

density of water = 1 g/cm^3;

dynamic viscosity of water = 0.01 g/(cm.s);

gravitational acceleration = 980 cm/s^2.

If the incoming water contains particles of diameter 25 \mu m(spherical and uniform), the removal efficiency of these particles is

51% | |

65% | |

78% | |

100% |

Question 8 Explanation:

\begin{aligned} Q&=0.2m^3/sec,\; L=32m\\ B&=8m,\; H=4m\\ \rho _p&=2.5gm/cc=2500kg/m^3\\ \rho _w&=1gm/cc=1000kg/m^3\\ \mu &=0.01gm/cm-sec=1 \times 10^{-3}kg/m-sec\\ d&=25\mu m=25 \times 10^{-6}m\\ g&=980cm/sec^2=9.8m/sec^2\\ v_0&=\frac{Q}{A_s}=\frac{Q}{L \times B}\\ &=\frac{0.2}{32 \times 8}=7.8125 \times 10^{-4}m/sec\\ v_s&=\frac{g(\rho _p-\rho _w)d^2}{18\mu }\\ &=\frac{9.8(2500-1000)\times (25 \times 10^{-6})^2}{18 \times (1 \times 10^{-3})}\\ &=5.1041 \times 10^{-4}m/sec\\ \eta &=\frac{v_s}{v_0}\times 100\\ &=\frac{5.1041 \times 10^{-4}}{7.8125 \times 10^{-4}}\times 100\\ &=65.33\%\simeq 65\% \end{aligned}

Question 9 |

A flocculation tank contains 1800\: m^{3} of water, which is mixed using paddles at an average velocity gradient G of 100/s. The water temperature and the corresponding dynamic viscosity are 30^{\circ}C\: \: and \: \: 0.798\times 10^{-3}\: \: Ns/m^{2} respectively. The theoretical power required to achieve the stated value of G (in kW, up to two decimal places) is ______

5.78 | |

10.25 | |

14.36 | |

20.65 |

Question 9 Explanation:

Power required

\begin{aligned} P &=\mu \mathrm{VG}^{2} \\ &=0.798 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} \times 1800 \mathrm{m}^{3} \times(1005)^{2} \\ &=14364 \mathrm{Nm} / \mathrm{s} \text { or } \mathrm{Watt} \\ &=14.364 \mathrm{kW} \\ &=14.36 \mathrm{kW} \end{aligned}

\begin{aligned} P &=\mu \mathrm{VG}^{2} \\ &=0.798 \times 10^{-3} \mathrm{Ns} / \mathrm{m}^{2} \times 1800 \mathrm{m}^{3} \times(1005)^{2} \\ &=14364 \mathrm{Nm} / \mathrm{s} \text { or } \mathrm{Watt} \\ &=14.364 \mathrm{kW} \\ &=14.36 \mathrm{kW} \end{aligned}

Question 10 |

A rapid sand filter comprising a number of filter beds is required to produce 99 MLD of potable water. Consider water loss during backwashing as 5%, rate of filtration as 6.0 m/h and length to width ratio of filter bed as 1.35. The width of each filter bed is to be kept equal to 5.2 m. One additional filter bed is to be provided to take care of break-down, repair and maintenance. The total number of filter beds required will be

19 | |

20 | |

21 | |

22 |

Question 10 Explanation:

Total water to be filtered

=99 \times 1.05 \mathrm{MLD}=103.95 \mathrm{MLD}

(Addition of 5 \% to be used for backwashing)

\begin{aligned} \frac{L}{B} &=1.35 \quad \text { where } B=5.2 \mathrm{m} \\ \therefore &=7.02 \mathrm{m} \end{aligned}

\therefore Surface area of each filter =36.504 \mathrm{m}^{2}

Total surface area required

\begin{array}{l} =\frac{\text { Discharge through filter }}{\text { Rate of filtration }} \\ =\frac{103.95 \times 10^{3}}{6 \times 24}=721.875 \mathrm{m}^{2} \end{array}

Total no. of working units required

\begin{array}{l} =\frac{721.875}{36.504} \\ =19.77 \text { filters }=20 \text { filters } \end{array}

1 unit is to added as standby, thus total no. of units required = 21

=99 \times 1.05 \mathrm{MLD}=103.95 \mathrm{MLD}

(Addition of 5 \% to be used for backwashing)

\begin{aligned} \frac{L}{B} &=1.35 \quad \text { where } B=5.2 \mathrm{m} \\ \therefore &=7.02 \mathrm{m} \end{aligned}

\therefore Surface area of each filter =36.504 \mathrm{m}^{2}

Total surface area required

\begin{array}{l} =\frac{\text { Discharge through filter }}{\text { Rate of filtration }} \\ =\frac{103.95 \times 10^{3}}{6 \times 24}=721.875 \mathrm{m}^{2} \end{array}

Total no. of working units required

\begin{array}{l} =\frac{721.875}{36.504} \\ =19.77 \text { filters }=20 \text { filters } \end{array}

1 unit is to added as standby, thus total no. of units required = 21

There are 10 questions to complete.