Past Year Questions: Air and Noise Pollution | Topic wise GATE Past Year Papers for Civil Engineering - Civil Engineering (CE) PDF Download

Clear all your Civil Engineering (CE) doubts with EduRev Join for Free

Join for Free

Q1: What is the CORRECT match between the air pollutants and treatment techniques given in the table? [2024, Set-2]
(a) P-i, Q-ii, R-iii, S-iv
(b) P-ii, Q-i, R-iv, S-iii
(c) P-ii, Q-iii, R-iv, S-i
(d) P-iv, Q-iii, R-i, S-ii
Ans: (d)
Sol: 
(I) Following treatment techniques are used to remove particulate Matter (PM).
Settling chambers
Inertial or Impact separators
Centrifugal separators or Cyclonic separators
Filters
(e) Electrostatic precipitators
(f) Scrubbers or wet collectors
(II) Following treatment techniques are used to remove/control gaseous contaminants
Combustion techniques
This method is used when gases are of organic nature.
equipment used in combustion are:
(i) vapor incinerators
(ii) after burners
(iii) Flares (process is called as flaring)
Note: Flaring (i.e. combustion method) is suitable for the removal of carbon monoxide because during combustion, carbon reacts with carbon monoxide to form CO ₂ .
Absorption
In this method, effluent gases are passed through absorbers (or scrubbers) which contain liquid absorbents that remove various gaseous pollutants.
Adsorption technique
In this method, the effluent gases are passed through adsorbers which contain solids of porous structure.

Q2: For assessing the compliance with the emissions standards of incineration plants, correction needs to be applied to the measured concentrations of air pollutants. The emission standard (based on  11% Oxygen) for HCI is 50mg/Nm³ and the measured concentrations of HCl and Oxygen in flue gas are 42mg/Nm³ and 13% respectively. Assuming 21% Oxygen in air, the CORRECT statement is: [2024, Set-1]
(a) No compliance, as the corrected HCl emission is greater than the emission standard.
(b) Compliance is there, as the corrected HCl emission is lesser than the emission standard
(c) Compliance is there, as there is no need to apply the correction since Oxygen is greater than 11% and  HCl emission is lesser than the emission standard.
(d) No compliance, as the Oxygen is greater than 11% in the flue gas.
Ans: (a)
Sol: 
As per CPCB, oxygen correction factor is calculated as shown below-
E_s = (21 - 08)E _m/21 - 0_m
Where
E_s = Calculated emission concentration in % at the standard percentage oxygen concentration in %
E _m = Measured emission concentration in %
O _s =Standard oxygen concentration in %
O _m = Measured oxygen concentration in % 
Note: This correction is done only when measured % oxygen concentration is higher than standard % oxygen (i. e. 11%)
So, E_s  = 21 -11/21 - 13 x 42  = 52.5Mg/Nm³
Hence, no compliance, the corrected HCl emission is greater than emission standard i.e. 52.5Mg/Nm³

Q3: The primary air pollutant(s) is/are    [2024, Set-1]
(a) Sulphur dioxide
(b) Lead
(c) Ozone
(d) Sulfuric acid
Ans: (b)
Sol: 
Lead is a particulate matter (PM) and its main source in atmosphere is automobiles.
SO₂ is first abundant atmospheric contaminant in many cities. It is produced by chemical interaction between Sulfuric and oxygen.

Q1: The theoretical aerobic oxidation of biomass (C5H7O2 N ) is given below:
C₅H₇O₂N + 5O₂ → 5CO₂ + NH₃ + 2H₂O
The biochemical oxidation of biomass is assumed as a first-order reaction with a rate constant of 0.23/d at 20°C(logarithm to base e). Neglecting the second-stage oxygen demand from its biochemical oxidation, the ratio of BOD5 at 2 0 ∘ C to total organic carbon (TOC) of biomass is (round off to two decimal places).
[Consider the atomic weights of C , H , O and N as 12 g/mol , 1 g/mol , 16 g/mol and 14 g/mol , respectively]    [2023, Set-2]
Ans: (1.8 to 2)
Sol: 
Calculation of BOD_s : 
C₅H₇O₂N + 5O₂ → 5CO₂ + NH₃ + 2H₂O
Considering 1 mole of biomass
BOD₄ = 160gm/mol
∴ BOD5 = BOD4 (1 - e ^{- kot} )
=160(1 − e ^{− 0.23 × 5})
=160 × 0.6833 = 109.34gm/mol
= 109.34gm/mol
Calculation of TOC: 
TOC = 12 × 5/113 × 113 = 60 g m / mol
 ( ∵ 1  mole biomass is considered)
∴ Required Ratio = 109.34/60 = 1.82

Q2: Match the following air pollutants with the most appropriate adverse health effects:    [2023, Set-2]

(a) (P) - (II), (Q) - (I), (R) - (IV), (S) - (III)
(b) (P) - (IV), (Q) - (I), (R) - (III), (S) - (II)
(c) (P) - (III), (Q) - (I), (R) - (II), (S) - (IV)
(d) (P) - (IV), (Q) - (I), (R) - (II), (S) - (III)
Ans: (c)
Sol: 
Aromatic hydrocarbons such as 3, 4benzpyrene, other polycyclic organic compounds which are originated due to incomplete combustion of hydrocarbons are considered to be carcinogenic agents. These are responsible for cancer.
Sulphur dioxide (SO₂) : It is an irritant gas which effects mucous membrane when inhaled. It leads to bronchial spasms. Asthma patients are badly affected.
Carbon Monoxide (CO) : Carbon monoxide has a strong affinity for combining with the oxyhemoglobin  of the blood to form  carboxyhemoglobin, C O H b. This reduces the ability of the oxyhemoglobin to carry oxygen to the body tissues. CO has about two hundred time the affinity of oxygen for attaching itself to the oxyhemoglobin, so that low levels of CO can still result in high levels of C O H b
Ozone ( O₃ ) can damage the tissue of the respiratory tract, causing inflammation and irritation, and result in symptoms such as coughing, chest tightness and worsening of asthma symptoms. In addition, ozone causes substantial damage to crops, forests and native plants.

Q3: For the elevation and temperature data given in the table, the existing lapse rate in the environment is °C/100 m (round off to two decimal places).    [2023, Set-2]
Ans: (0.84 to 0.85)
Sol: 
Lapse rate,
λ = Difference in height/Difference in Temperature  = ΔH/ΔT 
λ =  ΔH/ΔT=16.9−14.2/325−5
λ = 0.0084375 °C/m 
⇒ λ = 0.0084375 × 100/100 m 
⇒ λ=0.0084375/ 100 m = 0.84 °C/100m

Q4: Which of the following statements is/are TRUE in relation to the maximum mixing depth (or Height) ' D _max ' in the atmosphere? [2023, Set-1]
(a) D_max is always equal to the height of the layer of unstable air
(b) Ventilation coefficient depends on D_max
(c) A smaller D_max will have a smaller air pollution potential if other meteorological conditions remain same
(d) Vertical dispersion of pollutants occurs up to  D_{max 
Ans: (b, d) 
Sol:} 
The depth of mixing layer in which vertical movement of pollutants are possible is called maximum mixing depth (MMD).
An air parcel at temperature rises and cools. The level where its temperature becomes equal to surrounding air gives the maximum mixing depth value.
Also, ventilation coefficient = MMD × Average wind speed High value of ventilation coefficient leads to low air pollution potential.

Q1: A sample of air analyzed at 25  °C and 1 atm pressure is reported to contain 0.04 ppm of  SO₂  . Atomic mass of S = 32, O = 16. The equivalent SO₂  concentration (in μg/m³ ) will be__________. (round off to the nearest integer) [2022, Set-2]
Ans: (102 to 108)
Sol:
Concentration of SO₂ in ppm = 0.04 
Let's equivalent concentration in =μg/m³ is x. 
x μg of SO₂  present in 1m ³ of air. 
We know, 1 mole of SO ₂ (at 0 ° C , 1 atm) has volume of 22.4 L. 
at C 25 ° C and 1 atm, volume of SO ₂
22.4/273 + 10 x (273 + 25) = 24.45lit
X x 10 ^{- 6} /64  mole of SO₂ has volume
24.45 x X x10 ^-6/64 lit in 1m³ of air
0.382x × 10 ⁻³ m ³ of SO₂  present in 1 0 ⁶ m ³  of air. 
0.382 x × 1 0 ^{− 3} = 0.04 
0.382 x × 10 ⁻³ =0.04
x =104.7 ≈ 105 
0.04 ppm of SO₂ = 105 μg/m³ SO₂

Q2: A process equipment emits 5 kg/h of volatile organic compounds (VOCs). If a hood placed over the process equipment captures 95% of the VOCs, then the fugitive emission in kg/h is    [2022, Set-2]
(a) 0.25
(b) 4.75
(c) 2.5
(d) 0.48
Ans: (a)
Sol: 
VOC emission = 5 Kg/h 
Capturing Efficiency =95% 
Fugitive emission = (5/100) * 5 Kg/h = 0.25 Kg/h

Q1: Read the statements given below.
i. Value of the wind profile exponent for the 'very unstable' atmosphere is smaller than the wind profile exponent for the 'neutral' atmosphere.
ii. Downwind concentration of air pollutants due to an elevated point source will be inversely proportional to the wind speed.
iii. Value of the wind profile exponent for the 'neutral' atmosphere is smaller than the wind profile exponent for the 'very unstable' atmosphere.
iv. Downwind concentration of air pollutants due to an elevated point source will be directly proportional to the wind speed.
Select the correct option.    [2021, Set-2]
(a) i is False and ii is True
(b) i is True and iv is True
(c) ii is False and iii is False
(d) iii is False and iv is False
Ans: (d)
Sol: 
_{Concentration ∝ 1/wind speed}

Q2: Which of the following statement(s) is/are correct?    [2021, Set-2]
(a) Increased levels of carbon monoxide in the indoor environment result in the formation of carboxyhemoglobin and the long term exposure becomes a cause of cardiovascular diseases
(b) Volatile organic compounds act as one of the precursors to the formation of photochemical smog in the presence of sunlight
(c) Long term exposure to the increased level of photochemical smog becomes a cause of chest constriction and irritation of the mucous membrane
(d) Increased levels of volatile organic compounds in the indoor environment will result in the formation of photochemical smog which is a cause of cardiovascular diseases
Ans: (a)
Sol: Indoor environment photochemical smog does not form.

Q3: A baghouse filter has to treat 12 m³/s  of waste gas continuously. The baghouse is to be divided into 5 sections of equal cloth area such that one section can be shut down for cleaning and/or repairing, while the other 4 sections continue to operate. An air-to-cloth ratio of 6.0 m³/min − m² cloth will provide sufficient treatment to the gas. The individual bags are of 32 cm in diameter and 5 m in length. The total number of bags (in integer) required in the baghouse is ________________ [2021, Set-1]
Ans: (30 to 30)
Sol: 
Given, discharge to be passed = 12 m³/s 
 Out of 5 sections, 4 will operate at a time.
 Each bag is
Thus, surface area of each bag = π × 0.32 × 5 m² = 5.0265 m²
Total surface area required wr.t. given discharge

∴ Working bag filters required = 120m²/5.0265m² = 23.87 or 24 bag filters
But, since it is asked total (which includes a standby set also), then, Total no. of bag filters = 24 x 5/4 = 30

Q4: The liquid forms of particulate air pollutants are    [2021, Set-1]
(a) dust and mist
(b) mist and spray
(c)smoke and spray
(d) fly ash and fumes
Ans: (b)
Sol:
The liquid forms of particulate air pollutants are mist and spray.
Note: Mist is a cloud made of very small drops of water in the air just above the ground which reduces the visibility

Q5: Which one of the following statements is correct?    [2021, Set-1]
(a) Pyrolysis is an endothermic process, which takes place in the absence of oxygen
(b) Pyrolysis is an exothermic process, which takes place in the absence of oxygen
(c) Combustion is an endothermic process, which takes place in the abundance of oxygen
(d)Combustion is an exothermic process, which takes place in the absence of oxygen
Ans: (a)
Sol: 
Pyrolysis is an endothermic process as there is a substantial heat input required to raise the biomass to the reaction temperature.

Q1: A waste to energy plant burns dry solid waste of composition :
Carbon = 35%,
Oxygen = 26%,
Hydrogen = 10%,
Sulphur = 6%,
Nitrogen = 3% and
Inerts = 20%.
Burning rate is 1000 tonnes /day Oxygen in air by weight is 23%. Assume complete conversion of Carbon to CO₂  Hydrogen to H₂O Sulphur to SO₂ and Nitrogen to NO_2.
Given Atomic weighs : H=1, C=12, N=14, O=16, S=32.
The stoichiometric (theoretical) amount of air (in tonnes/day, round off to the nearest integer) required for complete burning of this waste, is __________.    [2020, Set-2]
Ans: (6963 to 6967)
Sol: 
Oxygen required for 350 tonne/day 
32/12 × 350 = 933.33 
Oxygen required for 100 tonne/day
32/4 × 100 = 800 Oxygen required for 30 tonne/day
32/14 x 30 = 68.57
Total O₂ = 1861.9 tonne/day 
Available O₂ in waste = 260 tonne/day 
Required =1861.9-260=1601.9 tonne /day
Amount of air required =1601.9/0.23 tonne/day ≃6965 tonne/day

Q2: A gas contains two types of suspended particle having average sizes of 2 μm and 50 μm. Amongst the options given below, the most suitable pollution control strategy for removal of these particles is    [2020, Set-2]
(a) settling chamber followed by bag filter
(b) electrostatic precipitator followed by venturi scrubber
(c)electrostatic precipitator followed by cyclonic separator
(d) bag filter followed by electrostatic precipitator
Ans: (a)
Sol:
Large size particles should be removed first, hence answer should be (A).
Theoretically, gravitational settling chamber should be able to remove particulates down to 5 or 10 μm, but in actual they are not practical for removal of particles much less than 50 μm in size.
Hence, gravitational settling chamber is most suitable choice for removal of suspended particles of size 50 μm.
After the removal of 50 μm size particles, to remove 2 μm size particles, we use fabric bag filter.
Note: Fabric bag filter has high collection efficiency for all particle sizes, especially for particles smaller than 10 μm in diameter.

Q3: A gaseous chemical has a concentration of 41.6 μmol/m ³ in air at 1 atm pressure and temperature 293 K. The universal gas constant R is 82.05 × 1 0 ^{− 6} ( m³ atm ) / ( mol K ) . Assuming that ideal gas law is valid, the concentration of the gaseous chemical (in ppm, round off to one decimal place), is _______. [2020, Set-1]
Ans: (0.9 to 1.1)
Sol: 
PV = nRT
V= nRT/P
= 10^-6 M³
41.6μ mole of gas volume of 1 0 ^{− 6} m³
41.6 μmol/m ³.