Question 1 |

In the context of Municipal Solid Waste Management, 'Haul' in 'Hauled Container System operated in conventional mode' includes the
____

time spent by the transport truck at the disposal site | |

time spent by the transport truck in traveling between a pickup point and the disposal site with a loaded container | |

time spent by the transport truck in picking up a loaded container at a pickup point | |

time spent by the transport truck in driving from the depot to the first pickup point |

Question 1 Explanation:

In hauled container system, 'Haul' time includes. The time required to reach the disposal site, starting after the container whose contents are to be emptied has been loaded on the track, plus the time after leaving the disposal site until the truck arrives at the location where the empty container is to be redeposited.

Time spent at the disposal site is not included.

Time spent at the disposal site is not included.

Question 2 |

The composition and energy content of a representative solid waste sample are given in the table. If the moisture content of the waste is 26 \%, the energy content of the solid waste on dry-weight basis is _____ \mathrm{MJ} / \mathrm{kg} (round off to one decimal place).

\begin{array}{|l|l|l|} \hline Component & \text{Percent} & \text{Energy} \\ & \text{by mass} & \text{content} \\ & & \text{as-discarded} \\ & &\text{basis} \\& & (\mathbf{M J} / \mathbf{k g}) \\ \hline Food \; waste & 20 & 4.5 \\ \hline Paper & 45 & 16.0 \\ \hline Cardboard & 5 & 14.0 \\ \hline Plastics & 10 & 32.0 \\ \hline Others & 20 & 8.0 \\ \hline \end{array}

\begin{array}{|l|l|l|} \hline Component & \text{Percent} & \text{Energy} \\ & \text{by mass} & \text{content} \\ & & \text{as-discarded} \\ & &\text{basis} \\& & (\mathbf{M J} / \mathbf{k g}) \\ \hline Food \; waste & 20 & 4.5 \\ \hline Paper & 45 & 16.0 \\ \hline Cardboard & 5 & 14.0 \\ \hline Plastics & 10 & 32.0 \\ \hline Others & 20 & 8.0 \\ \hline \end{array}

18.4 | |

28.2 | |

36.4 | |

12.2 |

Question 2 Explanation:

Assuming 1 \mathrm{Kg} of solid waste

\begin{array}{|c|c|c|c|c|} \hline \text {Component} & \text {Percent}& \text {Mas (Kg)} & \text { Energy} & \text { Energy}\\ & \text{by mass}& & \text{content} & \text{of waste (MJ)}\\ \hline Food \; waste & 20 & 0.2 & 4.5 & 0.2 \times 4.5=0.9 \\ \hline Paper & 45 & 0.45 & 16.0 & 0.45 \times 16=7.2 \\ \hline Carboard & 5 & 0.05 & 14.0 & 0.05 \times 14=0.7\\ \hline Plastics & 10 & 0.10 & 32.0 & 0.1 \times 32=32 \\ \hline Others & 20 & 0.20 & 80 & 0.2 \times 8=1.6 \\ \hline \end{array}

Energy produce from 1 \mathrm{Kg} work =13.6 \mathrm{~mJ}

Moisture content =26 \%

Solid content =100-26=74 \%

Therefore 1 \mathrm{Kg} of waste consist of =0.74 \mathrm{Kg} dry mass

\therefore \quad Energy content based on dry weight

\begin{aligned} & =\frac{13.6}{0.74}=18.378 \mathrm{~mJ} / \mathrm{Kg} \\ & \simeq 18.40 \mathrm{MJ} / \mathrm{Kg} \end{aligned}

\begin{array}{|c|c|c|c|c|} \hline \text {Component} & \text {Percent}& \text {Mas (Kg)} & \text { Energy} & \text { Energy}\\ & \text{by mass}& & \text{content} & \text{of waste (MJ)}\\ \hline Food \; waste & 20 & 0.2 & 4.5 & 0.2 \times 4.5=0.9 \\ \hline Paper & 45 & 0.45 & 16.0 & 0.45 \times 16=7.2 \\ \hline Carboard & 5 & 0.05 & 14.0 & 0.05 \times 14=0.7\\ \hline Plastics & 10 & 0.10 & 32.0 & 0.1 \times 32=32 \\ \hline Others & 20 & 0.20 & 80 & 0.2 \times 8=1.6 \\ \hline \end{array}

Energy produce from 1 \mathrm{Kg} work =13.6 \mathrm{~mJ}

Moisture content =26 \%

Solid content =100-26=74 \%

Therefore 1 \mathrm{Kg} of waste consist of =0.74 \mathrm{Kg} dry mass

\therefore \quad Energy content based on dry weight

\begin{aligned} & =\frac{13.6}{0.74}=18.378 \mathrm{~mJ} / \mathrm{Kg} \\ & \simeq 18.40 \mathrm{MJ} / \mathrm{Kg} \end{aligned}

Question 3 |

Which of the following statements is/are TRUE for
the Refuse-Derived Fuel (RDF) in the context of
Municipal Solid Waste (MSW) management?

Higher Heating Value (HHV) of the unprocessed
MSW is higher than the HHV of RDF processed
from the same MSW. | |

RDF can be made in the powdered form | |

Inorganic fraction of MSW is mostly converted
to RDF. | |

RDF cannot be used in conjunction with oil. |

Question 3 Explanation:

(A) is false. The Higher Heating Value (HHV) of Refuse-Derived Fuel (RDF) processed from Municipal Solid Waste (MSW) is typically higher than the HHV of unprocessed MSW. This is because RDF processing removes non-combustible materials from MSW and concentrates the combustible fraction.

(B) is true. RDF can be produced in various forms including pellets, fluff, and powdered form. The form of RDF depends on the specific process and intended end use.

(C) is partially true. RDF processing typically removes a significant portion of the inorganic fraction of MSW, but not all of it. The remaining inorganic fraction is typically mixed with the RDF product.

(D) is false. RDF can be used as a fuel in a variety of combustion systems, including those that burn oil. In fact, RDF is often used as a supplemental fuel in cement kilns and other industrial processes that also use oil as a fuel.

(B) is true. RDF can be produced in various forms including pellets, fluff, and powdered form. The form of RDF depends on the specific process and intended end use.

(C) is partially true. RDF processing typically removes a significant portion of the inorganic fraction of MSW, but not all of it. The remaining inorganic fraction is typically mixed with the RDF product.

(D) is false. RDF can be used as a fuel in a variety of combustion systems, including those that burn oil. In fact, RDF is often used as a supplemental fuel in cement kilns and other industrial processes that also use oil as a fuel.

Question 4 |

At a municipal waste handling facility, 30 metric ton mixture of food waste,
yard waste, and paper waste was available. The moisture content of this mixture
was found to be 10%. The ideal moisture content for composting this mixture is
50%. The amount of water to be added to this mixture to bring its moisture
content to the ideal condition is _______metric ton. (in integer)

12 | |

18 | |

24 | |

30 |

Question 4 Explanation:

Wt. of water initially present =frac{10}{100} \times 30=3mT

Wt. of solid =30-3=27mT

In ideal condition,

Wt. of solid =27mT

M/C=50%

Wt. of water =27mT

Amount of water needed to get ideal condition = 27 - 3 = 24 mT

Question 5 |

In a city, the chemical formula of biodegradable fraction of municipal solid waste
(MSW) is C_{100}H_{250}O_{80}N. The waste has to be treated by forced-aeration
composting process for which air requirement has to be estimated.

Assume oxygen in air (by weight) = 23 %, and density of air = 1.3 kg/m^3. Atomic mass: C = 12, H = 1, O = 16, N = 14.

C and H are oxidized completely whereas N is converted only into NH_3 during oxidation.

For oxidative degradation of 1 tonne of the waste, the required theoretical volume of air (in m^3/tonne) will be (round off to the nearest integer)

Assume oxygen in air (by weight) = 23 %, and density of air = 1.3 kg/m^3. Atomic mass: C = 12, H = 1, O = 16, N = 14.

C and H are oxidized completely whereas N is converted only into NH_3 during oxidation.

For oxidative degradation of 1 tonne of the waste, the required theoretical volume of air (in m^3/tonne) will be (round off to the nearest integer)

4749 | |

8025 | |

1418 | |

1092 |

Question 5 Explanation:

C_{100}H_{250}O_{80}N+xO_2\rightarrow 100CO_2+\frac{247}{2}H_2O+NH_3

\begin{aligned} 80+x \times 2&=200+\frac{247}{2}\\ x&=121.75 \end{aligned}

1 mole of MSW required 121.75 mole of oxygen No. of moles of MSW

=\frac{1 \; ton}{(100 \times 12+250+80 \times 16+14)g}

No. of moles of O2 required =121.75 \times \frac{1 \; tonne}{2744g}

\begin{aligned} \text{wt. of }O_2&= \frac{121.75}{2744}\times 32 \; tonne\\ \text{Mass of air }&= \frac{1.42}{0.23}\times 10^3 Kg\\ &=6.174 \times 10^3kg\\ \text{Density of air }&= 1.3 Kg/m^3\\ \text{Volume of air }&= \frac{6.174}{1.3}\times 10^3 m^3\\ =4749.2 m^3 &\text{per tonne of MSW} \end{aligned}

\begin{aligned} 80+x \times 2&=200+\frac{247}{2}\\ x&=121.75 \end{aligned}

1 mole of MSW required 121.75 mole of oxygen No. of moles of MSW

=\frac{1 \; ton}{(100 \times 12+250+80 \times 16+14)g}

No. of moles of O2 required =121.75 \times \frac{1 \; tonne}{2744g}

\begin{aligned} \text{wt. of }O_2&= \frac{121.75}{2744}\times 32 \; tonne\\ \text{Mass of air }&= \frac{1.42}{0.23}\times 10^3 Kg\\ &=6.174 \times 10^3kg\\ \text{Density of air }&= 1.3 Kg/m^3\\ \text{Volume of air }&= \frac{6.174}{1.3}\times 10^3 m^3\\ =4749.2 m^3 &\text{per tonne of MSW} \end{aligned}

There are 5 questions to complete.