Question 1 |

A city generates 40 \times 10^{6} kg of municipal solid waste (MSW) per year, out of which only 10% is recovered/recycled and the rest goes to landfill. The landfill has a single lift of 3 m height and is compacted to a density of 550 kg/m^{3}. If 80% of the landfill is assumed to be MSW, the landfill area (in m^{2}, up to one decimal place) required would be ______

25743.3 | |

36548.4 | |

27272.7 | |

26845.6 |

Question 1 Explanation:

Total weight generated by city

=40\times 10^{6}\: kg/year

Weight of MSW going into landfill

=0.9\times 40\times 10^{6}\: kg/year

=36\times 10^{6}\: kg/year

Compacted density

=550\: kg/m^{3}

Compacted volume of MSW

=\frac{36\times 10^{6}\: kg/year}{550\: kg/m^{3}}

=65454.5454\: m^{3}/year

Total landfill volume = Vol. of MSD + Vol. of cover

Given, Volume of MSW =0.8\times Total landfill volume

\therefore Vol. of cover =0.2\times Total landfill volume

\therefore Total landfill volume =\frac{65454.5454}{0.8}m^{3}/year =81818.18175m^{3}/year

Height of landfill =3\: m

\therefore Area of lanfill=\frac{81818.18175}{3} =27272.7\: m^{2}/year

=40\times 10^{6}\: kg/year

Weight of MSW going into landfill

=0.9\times 40\times 10^{6}\: kg/year

=36\times 10^{6}\: kg/year

Compacted density

=550\: kg/m^{3}

Compacted volume of MSW

=\frac{36\times 10^{6}\: kg/year}{550\: kg/m^{3}}

=65454.5454\: m^{3}/year

Total landfill volume = Vol. of MSD + Vol. of cover

Given, Volume of MSW =0.8\times Total landfill volume

\therefore Vol. of cover =0.2\times Total landfill volume

\therefore Total landfill volume =\frac{65454.5454}{0.8}m^{3}/year =81818.18175m^{3}/year

Height of landfill =3\: m

\therefore Area of lanfill=\frac{81818.18175}{3} =27272.7\: m^{2}/year

Question 2 |

The wastewater form a city, containing a high concentration of biodegradable organics, is beingsteadily discharged into a flowing river at a location S. If the rate of aeration of the river water islower than the rate of degradation of the organics, then the dissolved oxygen of the river water

is lowest at the locations S | |

is lowest at a point upstream of the location S | |

remains constant all along the length of the river | |

is lowest at a point downstream of the location S |

Question 2 Explanation:

As given rate of aeration is less than rate of degradation which decreases with time/distance, minimum DO is observed downstream of point of disposal 'S' where both rate of a reaction and degradation becomes equals.

Question 3 |

In a certain situation, waste water discharged into a river mixer with the river
water instantaneously and completely. Following is the data available :

Waste water:

DO = 2.00 mg/L

Discharge rate = 1.10 m^{3}/s

River water:

DO = 8.3 mg/L

Flow rate = 8.70 m^{3}/s

Temperature = 20^{\circ}C

Initial amount of DO in the mixture of waste and river shall be

Waste water:

DO = 2.00 mg/L

Discharge rate = 1.10 m^{3}/s

River water:

DO = 8.3 mg/L

Flow rate = 8.70 m^{3}/s

Temperature = 20^{\circ}C

Initial amount of DO in the mixture of waste and river shall be

5.3 mg/L | |

6.5 mg/L | |

7.6 mg/L | |

8.4 mg/L |

Question 3 Explanation:

\begin{aligned} DO_{mix} &=\frac{Q_{W}DO_{W}+Q_{R}DO_{R}}{Q_{W}+Q_{R}} \\ &=\frac{\left ( 1.10\times 2.00 \right )+\left ( 8.70\times 8.3 \right )}{1.10+8.70} \\ &=7.6\: mg/L \end{aligned}

Question 4 |

Match Group-I (Characteristics of sewage discharged into inland waters ) with Group-II (Allowable limit, mg/L).

P-2 Q-5 R-4 S-2 | |

P-4 Q-1 R-6 S-4 | |

P-3 Q-1 R-4 S-2 | |

P-2 Q-1 R-6 S-3 |

Question 4 Explanation:

There are 4 questions to complete.