Question 1 |

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 1 Explanation:

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

Question 2 |

Raw municipal solid waste (MSW) collected from a city contains 70% decomposable material that can be converted to methane. The water content of the decomposable material is 35%. An elemental analysis of the decomposable material yields the following mass percent.

C : H : O : N : other = 44 : 6 : 43 : 0.8 : 6.2

The methane production of the decomposable material is governed by the following stoichiometric relation

C_aH_bO_cN_d+nH_2O\rightarrow mCH_4+sCO_2+dNH_3

Given atomic weights: C=12, H=1, O=16, N=14. The mass of methane produced (in grams, round off to 1 decimal place) per kg of raw MSW will be ______

C : H : O : N : other = 44 : 6 : 43 : 0.8 : 6.2

The methane production of the decomposable material is governed by the following stoichiometric relation

C_aH_bO_cN_d+nH_2O\rightarrow mCH_4+sCO_2+dNH_3

Given atomic weights: C=12, H=1, O=16, N=14. The mass of methane produced (in grams, round off to 1 decimal place) per kg of raw MSW will be ______

112.8 | |

185.2 | |

137.6 | |

165.4 |

Question 2 Explanation:

Let 1 kg of row municipal solid waste (MSW)

Decomposable material = 70% of 1 kg = 0.7 kg

Water content = 35%

Weight of solids in MSW =\left(\frac{100-35}{100}\right)\times0.7=0.455\;\mathrm{kg}=455\;\mathrm{gms}

Given: C:H:O:N: other = 44 : 6: 43: 0.8:6.2\begin{array}{l}\mathrm{Weight}\;\mathrm{of}\;\mathrm C=\frac{(44)}{\left(44+6+43+0.8+6.2\right)}\times455=200.2\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm H=\frac{\left(6\right)}{\left(44+6+43+0.8+6.2\right)}\times455=27.3\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm O=\frac{(43)}{\left(44+6+43+0.8+6.2\right)}\times455=195.6\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm N=\frac{\left(0.8\right)}{\left(44+6+43+0.8+6.2\right)}\times455=3.64\;\mathrm{gms}\\\Rightarrow12\mathrm a=200.2\Rightarrow\mathrm a=16.68\\1\times\mathrm b=27.3\Rightarrow\mathrm b=27.3\\16\times\mathrm c=195.6\Rightarrow\mathrm c=12.225\\14\times\mathrm d=3.64\Rightarrow\mathrm d=0.26\\\mathrm{Now},\;\mathrm{balancing}\;\mathrm{the}\;\mathrm{given}\;\mathrm{stoichiometic}\;\mathrm r\times\mathrm n\\{\mathrm C}_{\mathrm a}{\mathrm H}_{\mathrm b}{\mathrm O}_{\mathrm C}{\mathrm N}_{\mathrm d}+{\mathrm{nH}}_2\mathrm O\rightarrow{\mathrm{mCH}}_4+{\mathrm{SCO}}_2+{\mathrm{dNn}}_3\\12\mathrm a=12\mathrm m+12\mathrm s\Rightarrow12\mathrm m+12\mathrm s=200.2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(12\mathrm a=200.2\right)\\\mathrm b+2\mathrm n=4\mathrm m+3\mathrm d\Rightarrow4\mathrm m-2\mathrm n=26.52\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(6=27.3\right)\\16\mathrm c+16\mathrm n=325\Rightarrow\mathrm n=25-12.225\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(16\mathrm c=195.6\right)\\\therefore\;\;\;\;\;4\mathrm m-2\left(2\mathrm s-12.225\right)=26.52\\\Rightarrow4\mathrm m-4\mathrm s=2.07\\\Rightarrow12\mathrm m-12\mathrm s=6.21\\\mathrm{Also},\;12\mathrm m+12\mathrm s=200.2\\\Rightarrow\mathrm m=8.6\\\therefore\;\;\mathrm{Mass}\;\mathrm{of}\;\mathrm{methane}\;\mathrm{produced} \\ =16\times\mathrm m=16\times8.6=137.6\;\mathrm{gms}\end{array}

Decomposable material = 70% of 1 kg = 0.7 kg

Water content = 35%

Weight of solids in MSW =\left(\frac{100-35}{100}\right)\times0.7=0.455\;\mathrm{kg}=455\;\mathrm{gms}

Given: C:H:O:N: other = 44 : 6: 43: 0.8:6.2\begin{array}{l}\mathrm{Weight}\;\mathrm{of}\;\mathrm C=\frac{(44)}{\left(44+6+43+0.8+6.2\right)}\times455=200.2\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm H=\frac{\left(6\right)}{\left(44+6+43+0.8+6.2\right)}\times455=27.3\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm O=\frac{(43)}{\left(44+6+43+0.8+6.2\right)}\times455=195.6\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm N=\frac{\left(0.8\right)}{\left(44+6+43+0.8+6.2\right)}\times455=3.64\;\mathrm{gms}\\\Rightarrow12\mathrm a=200.2\Rightarrow\mathrm a=16.68\\1\times\mathrm b=27.3\Rightarrow\mathrm b=27.3\\16\times\mathrm c=195.6\Rightarrow\mathrm c=12.225\\14\times\mathrm d=3.64\Rightarrow\mathrm d=0.26\\\mathrm{Now},\;\mathrm{balancing}\;\mathrm{the}\;\mathrm{given}\;\mathrm{stoichiometic}\;\mathrm r\times\mathrm n\\{\mathrm C}_{\mathrm a}{\mathrm H}_{\mathrm b}{\mathrm O}_{\mathrm C}{\mathrm N}_{\mathrm d}+{\mathrm{nH}}_2\mathrm O\rightarrow{\mathrm{mCH}}_4+{\mathrm{SCO}}_2+{\mathrm{dNn}}_3\\12\mathrm a=12\mathrm m+12\mathrm s\Rightarrow12\mathrm m+12\mathrm s=200.2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(12\mathrm a=200.2\right)\\\mathrm b+2\mathrm n=4\mathrm m+3\mathrm d\Rightarrow4\mathrm m-2\mathrm n=26.52\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(6=27.3\right)\\16\mathrm c+16\mathrm n=325\Rightarrow\mathrm n=25-12.225\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(16\mathrm c=195.6\right)\\\therefore\;\;\;\;\;4\mathrm m-2\left(2\mathrm s-12.225\right)=26.52\\\Rightarrow4\mathrm m-4\mathrm s=2.07\\\Rightarrow12\mathrm m-12\mathrm s=6.21\\\mathrm{Also},\;12\mathrm m+12\mathrm s=200.2\\\Rightarrow\mathrm m=8.6\\\therefore\;\;\mathrm{Mass}\;\mathrm{of}\;\mathrm{methane}\;\mathrm{produced} \\ =16\times\mathrm m=16\times8.6=137.6\;\mathrm{gms}\end{array}

Question 3 |

A coal containing 2% sulfur is burned completely to ash in a brick kiln at a rate of 30 kg/min. The sulfur content in the ash was found to be 6% of the initial amount of sulfur present in the coal fed to the brick kiln. The molecular weights of S, H and O are 32, 1 and 16 g/mole, respectively. The annual rate of sulfur dioxide (SO_{2}) emission from the kiln (in tonnes/year, up to two decimal places) is ______

240.25 | |

592.88 | |

540.45 | |

369.36 |

Question 3 Explanation:

Coal burned in one year

\begin{aligned} &=30\times 24\times 60\times 365 \\ &=1.5768\times 10^{7}\: \text{kg} \end{aligned}

Sulfur content=\frac{2}{100}\times 1.5768\times 10^{7}\times 10^{-6}

=315.36\; tonnes/year

Sulphur content in ash

\begin{aligned} &=\frac{6}{100}\times 315.36 \\ &= 18.92\; tonnes/year \end{aligned}

Sulphur converted to SO_{2}=315.36-18.92

=296.44\; tonnes/year

\; \; \; S\, +\, O_{2}\rightarrow SO_{2}

1 mole of S is present in 1 mole of SO_{2}

32 gm of S is present in 64 gm of SO_{2}

\therefore Rate of SO_{2} emission

=\frac{64}{32}\times 296.44=592.88\: tonnes/year

\begin{aligned} &=30\times 24\times 60\times 365 \\ &=1.5768\times 10^{7}\: \text{kg} \end{aligned}

Sulfur content=\frac{2}{100}\times 1.5768\times 10^{7}\times 10^{-6}

=315.36\; tonnes/year

Sulphur content in ash

\begin{aligned} &=\frac{6}{100}\times 315.36 \\ &= 18.92\; tonnes/year \end{aligned}

Sulphur converted to SO_{2}=315.36-18.92

=296.44\; tonnes/year

\; \; \; S\, +\, O_{2}\rightarrow SO_{2}

1 mole of S is present in 1 mole of SO_{2}

32 gm of S is present in 64 gm of SO_{2}

\therefore Rate of SO_{2} emission

=\frac{64}{32}\times 296.44=592.88\: tonnes/year

Question 4 |

In the figures, Group I represents the atmospheric temperature profiles (P, Q, R and S) and Group II represents dispersion of pollutants from a smoke stack (1, 2, 3 and 4). In the figures of Group I, the dashed line represents the dry adiabatic lapse rate, whereas the horizontal axis represents temperature and the vertical axis represents the altitude.

The correct match is

The correct match is

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

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

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

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

Question 5 |

There are 20,000 vehicles operating in a city with an average annual travel of 12,000 km per vehicle. The NO_x emission rate is 2.0 g/km per vehicle. The total annual release of NO_x will be

4,80,000 kg | |

4,800 kg | |

480 kg | |

48 kg |

Question 5 Explanation:

Total no. of kms. Travelled by all the vehicles

=20000\times 12000\: km=24\times 10^{7}\: km

Total NO_{x} emission

=2\: g/km\times 24\times 10^{7}\: km

=48\times 10^{7}\: g

=48\times 10^{4}\: kg

=20000\times 12000\: km=24\times 10^{7}\: km

Total NO_{x} emission

=2\: g/km\times 24\times 10^{7}\: km

=48\times 10^{7}\: g

=48\times 10^{4}\: kg

Question 6 |

The composition of a municipal solid waste sample is given below:

The difference between the energy content of the waste sample calculated on dry basis and asdiscarded basis (in kJ/kg) would be

The difference between the energy content of the waste sample calculated on dry basis and asdiscarded basis (in kJ/kg) would be

9657 | |

3870 | |

5850 | |

6543 |

Question 6 Explanation:

Energy on dry basis: It will be the total energy when whole mass is dry

In current situation only 60% mass is dry energy corresponding to 60% dry mass=5805 kJ/kg

Energy corresponding to 100% dry mass=\frac{5805}{60}\times 100=9675\: kJ/kg

So, energy as on dry basis=9675 kJ/kg

Energy based on as discharged basis=5805 kJ/kg

So, difference=9675-5805=3870 kJ/kg

Question 7 |

A landfill is to be designed to serve a population of 200000 for a period of 25 years. The solid waste (SW) generation is 2 kg/person/day. The density of the un-compacted SW is 100 kg/m^{3} and a compaction ratio of 4 is suggested. The ratio of compacted fill (i.e. SW + cover) to compacted SW is 1.5. The landfill volume (in million m^{3} ) required is _______.

20.4 | |

13.6 | |

21.9 | |

15.6 |

Question 7 Explanation:

Solid waste generated per day

=2\times 200000=400000\: kg

Total solid waste generated in 25 years

=400000\times 365\times 25=3.65\times 10^{9}\: kg

Compaction ratio,

4=\frac{Volume\; before\; compaction}{Volume\; after\; compaction}

Volume of waste before compaction,

V=\frac{3.65\times 10^{9}}{100}=3.65\times 10^{7}\: m^{3}

Volume of waste after compaction,

V'=\frac{3.65\times 10^{7}}{4}=9.125\times 10^{6}\: m^{3}

\frac{SW+cover}{SW}=\frac{SW}{SW}+\frac{cover}{SW}=1.5

\frac{Cover}{SW}=0.5

Cover=0.5\times 9.125\times 10^{6}

\; \; \; =4.5625\times 10^{6}\: m^{3}

\begin{aligned} \text{Total volume} &=SW+Cover \\ &=\left ( 9.125+4.5625 \right )\times 10^{6}\: m^{3} \\ &=13.6875\: million\: m^{3} \end{aligned}

=2\times 200000=400000\: kg

Total solid waste generated in 25 years

=400000\times 365\times 25=3.65\times 10^{9}\: kg

Compaction ratio,

4=\frac{Volume\; before\; compaction}{Volume\; after\; compaction}

Volume of waste before compaction,

V=\frac{3.65\times 10^{9}}{100}=3.65\times 10^{7}\: m^{3}

Volume of waste after compaction,

V'=\frac{3.65\times 10^{7}}{4}=9.125\times 10^{6}\: m^{3}

\frac{SW+cover}{SW}=\frac{SW}{SW}+\frac{cover}{SW}=1.5

\frac{Cover}{SW}=0.5

Cover=0.5\times 9.125\times 10^{6}

\; \; \; =4.5625\times 10^{6}\: m^{3}

\begin{aligned} \text{Total volume} &=SW+Cover \\ &=\left ( 9.125+4.5625 \right )\times 10^{6}\: m^{3} \\ &=13.6875\: million\: m^{3} \end{aligned}

Question 8 |

Solid waste generated from an industry contains only two components, X and Y as shown in the table below

Assuming (c_1+c_2)=100, the composite density of the solid waste (\rho ) is given by :

Assuming (c_1+c_2)=100, the composite density of the solid waste (\rho ) is given by :

\frac{100}{\left ( \frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}} \right )} | |

100{\left ( \frac{\rho _{1}}{c_{1}}+\frac{\rho _{2}}{c_{2}} \right )} | |

100{\left ({c_{1}}{\rho _{1}}+{c_{2}}{\rho _{2}} \right )} | |

100\left ( \frac{\rho_{1} \rho_{2}}{c_{1}\rho_{1}+c_{2}\rho_2{}} \right ) |

Question 8 Explanation:

Let density of sludge is \rho ,

\therefore \; \; \frac{c_{1}+c_{2}}{\rho }=\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}

Assuming c_{1}+c_{2}=100

\therefore \; \; \rho =\frac{100}{\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}}

\therefore \; \; \frac{c_{1}+c_{2}}{\rho }=\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}

Assuming c_{1}+c_{2}=100

\therefore \; \; \rho =\frac{100}{\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}}

Question 9 |

A coastal city produces municipal solid waste (MSW) with high moisture content,
high organic materials, low calorific value and low inorganic materials. The most
effective and sustainable option for MSW management in that city is

Composting | |

Dumping is sea | |

Incineration | |

Landfill |

Question 9 Explanation:

Incineration can be adopted when the calorific value of the MSW is high. Land fill can be adopted when the density of MSW is high. Since in the given MSW the quality of inorganic material is low, its density is less. Composting can be adopted when the MSW contains high organic content. Barging the MSW into the sea is now a days generaly not used and has become obsolete.

Question 10 |

50 g of CO_{2} and 25 g of CH_{4} are produced from the decomposition of municipal solid waste (MSW) with a formula weight of 120 g. What is the average per capita green house gas production in a city of 1 million people with a MSW production rate of 500 ton/day ?

104 g/day | |

120 g/day | |

208 g/day | |

313 g/day |

Question 10 Explanation:

We know that both CO_{2} and CH_{4} are green house gases.

\therefore Total green house gases produced by 120 gm of MSW

= 50 g CO_{2} +25 g of CH_{4}=75 gm

500 ton of MSW will produce green house gases

\begin{aligned} &=500\times 10^{6}\times \frac{75}{120} \\ &= 312.5\times 10^{6}\: g/day \end{aligned}

Per capita average production of green house gases

\begin{aligned} &=\frac{312.5\times 10^{6}}{1\times 10^{6}}\: g/day \\ &=312.5\: g/day\approx 313\: g/day \end{aligned}

\therefore Total green house gases produced by 120 gm of MSW

= 50 g CO_{2} +25 g of CH_{4}=75 gm

500 ton of MSW will produce green house gases

\begin{aligned} &=500\times 10^{6}\times \frac{75}{120} \\ &= 312.5\times 10^{6}\: g/day \end{aligned}

Per capita average production of green house gases

\begin{aligned} &=\frac{312.5\times 10^{6}}{1\times 10^{6}}\: g/day \\ &=312.5\: g/day\approx 313\: g/day \end{aligned}

There are 10 questions to complete.