Question 1 |

A grit chamber of rectangular cross-section is to be designed to remove particles with diameter of 0.25 mm and specific gravity of 2.70. The terminal settling velocity of the particles is estimated as 2.5 cm/s. The chamber is having a width of 0.50 m and has to carry a peak wastewater flow of 9720 \mathrm{~m}^{3} / \mathrm{d} giving the depth of flow as 0.75 m. If a flow-through velocity of 0.3 m/s has to be maintained using a proportional weir at the outlet end of the chamber, the minimum length of the chamber (in m,in integer) to remove 0.25 mm particles completely is _________

5 | |

7 | |

9 | |

10 |

Question 1 Explanation:

Minimum length of chamber

\begin{aligned} L_{\min } &=V_{\text {flow }} \times t_{d} \\ V_{\text {fow }} &=0.30 \mathrm{~m} / \mathrm{s} \\ t_{d} &=\frac{H}{V_{\text {settling }}}=\frac{0.75}{2.5 \times 10^{-2}}=30 \mathrm{sec} \\ L_{\text {min }} &=0.3 \mathrm{~m} / \mathrm{sec} \times 30 \mathrm{sec} \\ &=9 \mathrm{~m} \end{aligned}

\begin{aligned} L_{\min } &=V_{\text {flow }} \times t_{d} \\ V_{\text {fow }} &=0.30 \mathrm{~m} / \mathrm{s} \\ t_{d} &=\frac{H}{V_{\text {settling }}}=\frac{0.75}{2.5 \times 10^{-2}}=30 \mathrm{sec} \\ L_{\text {min }} &=0.3 \mathrm{~m} / \mathrm{sec} \times 30 \mathrm{sec} \\ &=9 \mathrm{~m} \end{aligned}

Question 2 |

An activated sludge process (ASP) is designed for secondary treatment of 7500 m^{3} / day of municipal wastewater. After primary clarifier, the ultimate BOD of the influent, which enters into ASP reactor is 200 mg/L. Treated effluent after secondary clarifier is required to have an ultimate BOD of 20 mg/L. Mix liquor volatile suspended solids (MLVSS) concentration in the reactor and the underflow is maintained as 3000 mg/L and 12000 mg/L, respectively. The hydraulic retention time and mean cell residence time are 0.2 day and 10 days, respectively. A representative flow diagram of the ASP is shown below.

The underflow volume (in m^{3} / day, round off to one decimal place) of sludge wastage is _________

The underflow volume (in m^{3} / day, round off to one decimal place) of sludge wastage is _________

25.4 | |

37.5 | |

65.2 | |

87.6 |

Question 2 Explanation:

\begin{aligned} \text{Given,} \qquad X &=3000 \mathrm{mg} / l, \quad X_{\mathrm{u}}=12000 \mathrm{mg} / l \\ \text{Since} \qquad \mathrm{HRT} &=\frac{V}{Q_{0}}\\ \Rightarrow &\text{Volume of reactor,}\\ V &=Q_{0} \times H R T \\ &=7500 \times 0.2 \mathrm{~m}^{3} \\ &=1500 \mathrm{~m}^{3} \\ &\text{Sludge age,} \\ \theta_{\mathrm{C}} &=\frac{V X}{\left(Q_{0}-Q_{w}\right) X_{e}+Q_{w} X_{u}} \qquad \left(X_{e} \simeq 0\right)\\ 10&=\frac{1500 \times 3000}{Q_{W} \times 12000}\\ Q_{w}&=37.5 \mathrm{~m}^{3} / day \end{aligned}

Question 3 |

A water filtration unit is made of uniform-size sand particles of 0.4 mm diameter with a shape factor of 0.84 and specific gravity of 2.55. The depth of the filter bed is 0.70 m and the porosity is 0.35. The filter bed is to be expanded to a porosity of 0.65 by hydraulic backwash. In the terminal settling velocity of sand particles during backwash is 4.5 cm/s, the required backwash velocity is

5.79 \times 10^{-3} \mathrm{~m} / \mathrm{s} | |

6.35 \times 10^{-3} \mathrm{~m} / \mathrm{s} | |

0.69 cm/s | |

0.75 cm/s |

Question 3 Explanation:

\begin{aligned} n^{\prime} &=\text { Porosity of expanded bed } \\ n^{\prime} &=\left(\frac{V_{B}}{V_{s}}\right)^{0.22} \\ 0.65 &=\left(\frac{V_{B}}{4.5 \mathrm{~cm} / \mathrm{s}}\right)^{0.22} \\ V_{B} &=6.35 \times 10^{-3} \mathrm{~m} / \mathrm{s} \end{aligned}

Question 4 |

Read the statements given below.

i. Value of the wind profile exponent for the 'very unstable' atmosphere is smaller than the wind profile exponent for the 'neutral' atmosphere.

ii. Downwind concentration of air pollutants due to an elevated point source will be inversely proportional to the wind speed.

iii. Value of the wind profile exponent for the 'neutral' atmosphere is smaller than the wind profile exponent for the 'very unstable' atmosphere.

iv. Downwind concentration of air pollutants due to an elevated point source will be directly proportional to the wind speed.

Select the correct option.

i. Value of the wind profile exponent for the 'very unstable' atmosphere is smaller than the wind profile exponent for the 'neutral' atmosphere.

ii. Downwind concentration of air pollutants due to an elevated point source will be inversely proportional to the wind speed.

iii. Value of the wind profile exponent for the 'neutral' atmosphere is smaller than the wind profile exponent for the 'very unstable' atmosphere.

iv. Downwind concentration of air pollutants due to an elevated point source will be directly proportional to the wind speed.

Select the correct option.

i is False and ii is True | |

i is True and iv is True | |

ii is False and iii is False | |

iii is False and iv is False |

Question 4 Explanation:

\text{Concentration} \propto \frac{1}{\text{wind speed}}

Question 5 |

A lake has a maximum depth of 60 m. If the mean atmospheric pressure in the lake region is 91 kPa and the unit weight of the lake water is 9790 \mathrm{~N} / \mathrm{m}^{3}, the absolute pressure (in kPa,round off to two decimal places) at the maximum depth of the lake is ___________

678.4 | |

258.6 | |

458.2 | |

125.9 |

Question 5 Explanation:

Absolute pressure at maximum depth of the lake =P_{\text {atm }}+\rho g h

=91+\frac{9790(60)}{1000}=678.4 \mathrm{kPa}

=91+\frac{9790(60)}{1000}=678.4 \mathrm{kPa}

Question 6 |

The internal \left(d_{i}\right) and external \left(d_{o}\right) diameters of a Shelby sampler are 48 mm and 52 mm, respectively. The area ratio \left(A_{r}\right)
of the sampler (in %,round off to two decimal palces) is _____________

12.45 | |

28.56 | |

47.12 | |

17.36 |

Question 6 Explanation:

Outside diameter = 52 mm

Inside diameter= 48 mm

\begin{aligned} A_{r}&=\frac{\frac{\pi}{4}\left(D_{2}\right)^{2}-\frac{\pi}{4}\left(D_{1}\right)^{2}}{\frac{\pi}{4}\left(D_{1}\right)^{2}} \times 100 \\ A_{r}&=\frac{(52)^{2}-(48)^{2}}{(48)^{2}} \times 100=17.36 \% \end{aligned}

Inside diameter= 48 mm

\begin{aligned} A_{r}&=\frac{\frac{\pi}{4}\left(D_{2}\right)^{2}-\frac{\pi}{4}\left(D_{1}\right)^{2}}{\frac{\pi}{4}\left(D_{1}\right)^{2}} \times 100 \\ A_{r}&=\frac{(52)^{2}-(48)^{2}}{(48)^{2}} \times 100=17.36 \% \end{aligned}

Question 7 |

Which of the following statement(s) is/are correct?

Increased levels of carbon monoxide in the indoor environment result in the formation of carboxyhemoglobin and the long term exposure becomes a cause of cardiovascular diseases | |

Volatile organic compounds act as one of the precursors to the formation of photochemical smog in the presence of sunlight | |

Long term exposure to the increased level of photochemical smog becomes a cause of chest constriction and irritation of the mucous membrane | |

Increased levels of volatile organic compounds in the indoor environment will result in the formation of photochemical smog which is a cause of cardiovascular diseases |

Question 7 Explanation:

Indoor environment photochemical smog does not form.

Question 8 |

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 8 Explanation:

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

Question 9 |

Determine the correctness or otherwise of the following Assertion [a] and the Reason [r].

Assertion [a]: One of the best ways to reduce the amount of solid wastes is to reduce the consumption of raw materials.

Reason [r]: Solid wastes are seldom generated when raw materials are converted to goods for consumption.

Assertion [a]: One of the best ways to reduce the amount of solid wastes is to reduce the consumption of raw materials.

Reason [r]: Solid wastes are seldom generated when raw materials are converted to goods for consumption.

Both [a] and [r] are true and [r] the correct reason for [a] | |

Both [a] and [r] are true but [r] is not the correct reason for [a] | |

Both [a] and [r] are false | |

[a] is true but [r] is false |

Question 9 Explanation:

Solid waste generated by the use of raw material, processing and finished goods

Question 10 |

A secondary clarifier handles a total flow of 9600 \mathrm{~m}^{3} / \mathrm{d} from the aeration tank of a conventional activated-sludge treatment system. The concentration of solids in the flow from the aeration tank is 3000 mg/L. The clarifier is required to thicken the solids to 12000 mg/L, and hence it is to be designed for a solid flux of 3.2 \frac{\mathrm{kg}}{\mathrm{m}^{2} \cdot \mathrm{h}}. The surface area of the designed clarifier for thickening (in \mathrm{~m}^{2}, in integer) is _________________

125 | |

258 | |

525 | |

375 |

Question 10 Explanation:

\begin{aligned} Q &=9600 \mathrm{~m}^{3} / \mathrm{d} \\ X &=3000 \mathrm{mg} / l \\ X_{u} &=12000 \mathrm{mg} / l \\ \text {Solid flux }& =3.2 \mathrm{~kg} / \mathrm{m}^{2}-\mathrm{h}\\ \text{Surface area,} \qquad \mathrm{A}&=\frac{\text { Total quantity of solids entering }}{\text { Solid flux }}\\ A &=\frac{9600 \mathrm{~m}^{3} / \mathrm{d} \times 3000 \mathrm{mg} / l}{3.2 \times 10^{6} \mathrm{mg} / \mathrm{m}^{2}-\mathrm{h}} \\ &=\frac{9600 \times 10^{3} \mathrm{lld} \times 3000 \mathrm{mg} / \mathrm{l}}{3.2 \times 10^{6} \mathrm{mg} / \mathrm{m}^{2}-\mathrm{h} \times 24 \mathrm{~h} / \mathrm{d}} \\ &=375 \mathrm{~m}^{2} \end{aligned}

There are 10 questions to complete.

Question 49 of environmental engineering is provided with flawed data

Thank You Kishan Agarwal,

We have updated the figure in question.