Municipal Solid Waste Management

Question 1
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)
GATE CE 2022 SET-2   Environmental Engineering
Question 1 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
Wt. of water =27mT
Amount of water needed to get ideal condition = 27 - 3 = 24 mT
Question 2
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)
GATE CE 2022 SET-2   Environmental Engineering
Question 2 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}
Question 3
To design an optimum municipal solid waste collection route, which of the following is/are NOT desired:
Collection vehicle should not travel twice down the same street in a day.
Waste collection on congested roads should not occur during rush hours in morning or evening.
Collection should occur in the uphill direction.
The last collection point on a route should be as close as possible to the waste disposal facility.
GATE CE 2022 SET-2   Environmental Engineering
Question 3 Explanation: 
Some heuristic guidelines that should be taken into consideration when laying out routes are as follows :
1. Existing policies and regulations related to such items as the point of collection and frequency of collection must be identified.
2. Existing system characteristics such as crew size and vehicle types must be coordinated.
3. Wherever possible, routes should be laid out so that they begin and end near arterial streets, using topographical and physical barriers as route boundaries.
4. In hilly area, routes should start at the top of the grade and proceed downhill as the vehicle becomes loaded.
5. Routes should be laid out so that the last container to be collected on the route is located nearest to the disposal site.
6. Waste generated at traffic-congested locations should be collected as early in the day as possible.
7. Sources at which extremely large quantities of waste are generated should be serviced during the first part of the days.
8. Scattered pickup points (where small quantities of solid waste are generated) that receive the same collection frequency should, if possible, be serviced during one trip or on the same day.
Question 4
In a solid waste handling facility, the moisture contents (MC) of food waste, paper waste, and glass waste were found to be MC_f, MC_p, and MC_g, respectively. Similarly, the energy contents (EC) of plastic waste, food waste, and glass waste were found to be EC_{pp}, EC_f, and EC_g, respectively. Which of the following statement(s) is/are correct?
MC_f \gt MC_p \gt MC_g
EC_{pp} \gt EC_f \gt EC_g
MC_f \lt MC_p \lt MC_g
EC_{pp} \lt EC_f \lt EC_g
GATE CE 2022 SET-2   Environmental Engineering
Question 4 Explanation: 
Typical speicifiv weight and moisture content data for residential, commercial, industraial, and agricultural wastes.

Moisture content
Mcf > Mcp > Mcg
Typical values for short residue and energy conent of residentail MSW.

Energy content
ECpp > ECf > ECg
Question 5
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.
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
GATE CE 2021 SET-2   Environmental Engineering
Question 5 Explanation: 
Solid waste generated by the use of raw material, processing and finished goods
Question 6
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 ______
GATE CE 2019 SET-2   Environmental Engineering
Question 6 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 7
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 ______
GATE CE 2018 SET-2   Environmental Engineering
Question 7 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 8
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
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
GATE CE 2018 SET-2   Environmental Engineering
Question 9
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
GATE CE 2018 SET-1   Environmental Engineering
Question 9 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 10
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
GATE CE 2017 SET-2   Environmental Engineering
Question 10 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
There are 10 questions to complete.

Leave a Comment