# Municipal Solid Waste Management

 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.
 A Both [a] and [r] are true and [r] the correct reason for [a] B Both [a] and [r] are true but [r] is not the correct reason for [a] C Both [a] and [r] are false D [a] is true but [r] is false
GATE CE 2021 SET-2   Environmental Engineering
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 ______
 A 112.8 B 185.2 C 137.6 D 165.4
GATE CE 2019 SET-2   Environmental Engineering
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$\begin{array}{l}\mathrm{Weight}\;\mathrm{of}\;\mathrm C=\frac{(44)}{\left(44+6+43+0.8+6.2\right)}\times455=200.2\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm H=\frac{\left(6\right)}{\left(44+6+43+0.8+6.2\right)}\times455=27.3\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm O=\frac{(43)}{\left(44+6+43+0.8+6.2\right)}\times455=195.6\;\mathrm{gms}\\\mathrm{Weight}\;\mathrm{of}\;\mathrm N=\frac{\left(0.8\right)}{\left(44+6+43+0.8+6.2\right)}\times455=3.64\;\mathrm{gms}\\\Rightarrow12\mathrm a=200.2\Rightarrow\mathrm a=16.68\\1\times\mathrm b=27.3\Rightarrow\mathrm b=27.3\\16\times\mathrm c=195.6\Rightarrow\mathrm c=12.225\\14\times\mathrm d=3.64\Rightarrow\mathrm d=0.26\\\mathrm{Now},\;\mathrm{balancing}\;\mathrm{the}\;\mathrm{given}\;\mathrm{stoichiometic}\;\mathrm r\times\mathrm n\\{\mathrm C}_{\mathrm a}{\mathrm H}_{\mathrm b}{\mathrm O}_{\mathrm C}{\mathrm N}_{\mathrm d}+{\mathrm{nH}}_2\mathrm O\rightarrow{\mathrm{mCH}}_4+{\mathrm{SCO}}_2+{\mathrm{dNn}}_3\\12\mathrm a=12\mathrm m+12\mathrm s\Rightarrow12\mathrm m+12\mathrm s=200.2\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(12\mathrm a=200.2\right)\\\mathrm b+2\mathrm n=4\mathrm m+3\mathrm d\Rightarrow4\mathrm m-2\mathrm n=26.52\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(6=27.3\right)\\16\mathrm c+16\mathrm n=325\Rightarrow\mathrm n=25-12.225\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\left(16\mathrm c=195.6\right)\\\therefore\;\;\;\;\;4\mathrm m-2\left(2\mathrm s-12.225\right)=26.52\\\Rightarrow4\mathrm m-4\mathrm s=2.07\\\Rightarrow12\mathrm m-12\mathrm s=6.21\\\mathrm{Also},\;12\mathrm m+12\mathrm s=200.2\\\Rightarrow\mathrm m=8.6\\\therefore\;\;\mathrm{Mass}\;\mathrm{of}\;\mathrm{methane}\;\mathrm{produced} \\ =16\times\mathrm m=16\times8.6=137.6\;\mathrm{gms}\end{array}$
 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 ______
 A 240.25 B 592.88 C 540.45 D 369.36
GATE CE 2018 SET-2   Environmental Engineering
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
$=\frac{64}{32}\times 296.44=592.88\: tonnes/year$
 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
 A P-1, Q-2, R-3, S-4 B P-1, Q-2, R-4, S-3 C P-1, Q-4, R-3, S-2 D P-3, Q-1, R-2, S-4
GATE CE 2018 SET-2   Environmental Engineering
 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
 A 4,80,000 kg B 4,800 kg C 480 kg D 48 kg
GATE CE 2018 SET-1   Environmental Engineering
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
$=2\: g/km\times 24\times 10^{7}\: km$
$=48\times 10^{7}\: g$
$=48\times 10^{4}\: kg$
 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
 A 9657 B 3870 C 5850 D 6543
GATE CE 2017 SET-2   Environmental Engineering
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 _______.
 A 20.4 B 13.6 C 21.9 D 15.6
GATE CE 2015 SET-2   Environmental Engineering
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}
 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 :
 A $\frac{100}{\left ( \frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}} \right )}$ B $100{\left ( \frac{\rho _{1}}{c_{1}}+\frac{\rho _{2}}{c_{2}} \right )}$ C $100{\left ({c_{1}}{\rho _{1}}+{c_{2}}{\rho _{2}} \right )}$ D $100\left ( \frac{\rho_{1} \rho_{2}}{c_{1}\rho_{1}+c_{2}\rho_2{}} \right )$
GATE CE 2015 SET-1   Environmental Engineering
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 $c_{1}+c_{2}=100$
$\therefore \; \; \rho =\frac{100}{\frac{c_{1}}{\rho _{1}}+\frac{c_{2}}{\rho _{2}}}$
 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
 A Composting B Dumping is sea C Incineration D Landfill
GATE CE 2010   Environmental Engineering
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 ?
 A 104 g/day B 120 g/day C 208 g/day D 313 g/day
GATE CE 2007   Environmental Engineering
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}
There are 10 questions to complete.