Air And Noise Pollution Practice Questions

To find the minimum size of the particle removed with 100% efficiency, we equate the settling velocity of the particle to the vertical distance the particle must settle during the residence time in the chamber.Residence time in the chamber is: $t = \frac{\text{length of chamber}}{\text{horizontal velocity}} = \frac{12}{0.2} = 60\ \text{seconds}$ So, for 100% removal, the particle must fall the height of the chamber in 60 seconds: $v_s = \frac{\text{height}}{t} = \frac{2}{60} = 0.0333\ \text{m/s}$ Using Stokes' Law (valid under laminar conditions): $v_s = \frac{g d^2 (\rho_p - \rho_f)}{18 \mu}$ Where: $d = \text{diameter of particle (\in m)}$ $\rho_p = \text{density of particle} = 2.65 \times 1000 = 2650\ \text{kg/m}^3$ $\rho_f = \text{density of air} = 1.0\ \text{kg/m}^3$ $\mu = \text{viscosity of air} = 2.1 \times 10^{-5}\ \text{kg/m?s}$ Now solve for $d$ : $0.0333 = \frac{9.81 \cdot d^2 (2650 - 1)}{18 \cdot 2.1 \times 10^{-5}}$ $0.0333 = \frac{9.81 \cdot d^2 \cdot 2649}{3.78 \times 10^{-4}}$ $d^2 = \frac{0.0333 \cdot 3.78 \times 10^{-4}}{9.81 \cdot 2649}$ $d^2 \approx \frac{1.25874 \times 10^{-5}}{25983.69} \approx 4.843 \times 10^{-10}$ $d \approx \sqrt{4.843 \times 10^{-10}} \approx 2.2 \times 10^{-5}\ \text{m}$ Convert to micrometers: $d = 2.2 \times 10^{-5} \times 10^6 = 22.0\ \mu m$

Lead is a particulate matter (PM) and its main source in atmosphere is automobiles. $SO_2$ is first abundant atmospheric contaminant in many cities. It is produced by chemical interaction between sulphur and oxygen.

(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. Equipments used in combustion are: (i) vapour 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 $\mathrm{CO}_{2}$ . 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.

As per $\mathrm{CPCB}$ , oxygen correction factor is calculated as shown below- $E_{s}=\frac{\left(21-0_{s}\right) E_{m}}{21-0_{m}}$ Where $E_{s}=$ Calculated emission concentration in $\%$ at the standard percentage oxygen concentration in % $E_{m}=$ Measured emission concentration in % $\mathrm{O}_{s}=$ Standard oxygen concentration in % $\mathrm{O}_{m}=$ Measured oxygen concentration in % Note: This correction is done only when measured % oxygen concentration is higher than standard % oxygen (i.e. $11 \%$ ) So, $E_{s}=\frac{21-11}{21-13} \times 42=52.5 \mathrm{Mg} / \mathrm{Nm}^{3}$ Hence, no compliance, the corrected $\mathrm{HCl}$ emission is greater than emission standard i.e. $50 \mathrm{mg} / \mathrm{Nm}^{3}$ .

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 $\left(\mathrm{SO}_{2}\right)$ : 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 haemoglobin of the blood to form carboxyhemoglobin, $\mathrm{COHb}$ . This reduces the ability of the haemoglobin to carry oxygen to the body tissues. CO has about two hundred time the affinity of oxygen for attaching itself to the haemoglobin, so that low levels of CO can still result in high levels of $\mathrm{COHb}$ . Carbon monoxide also affects the central nervous system. Ozone $\left(\mathrm{O}_{3}\right)$ 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.

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 $\times$ Average wind speed High value of ventilation coefficient leads to low air pollution potential.

Lapse rate, $\lambda=\frac{\text { Difference \in Temperature }}{\text { Difference \in height }}=\frac{\Delta \mathrm{T}}{\Delta \mathrm{H}}$ $\lambda=\frac{\Delta \mathrm{T}}{\Delta \mathrm{H}}=\frac{16.9-14.2}{325-5}$ $\lambda=0.0084375^{\circ} \mathrm{C} / \mathrm{m}$ $\Rightarrow \quad \lambda=\frac{0.0084375 \times 100}{100 \mathrm{~m}}$ $\Rightarrow \quad \lambda=\frac{0.084375^{\circ} \mathrm{C}}{100 \mathrm{~m}}=0.84^{\circ} \mathrm{C} / 100 \mathrm{~m}$

Calculation of $B O D_{s}$ : $\mathrm{C}_{5} \mathrm{H}_{7} \mathrm{O}_{2} \mathrm{~N}+5 \mathrm{O}_{2} \rightarrow 5 \mathrm{CO}_{2}+\mathrm{NH}_{3}+2 \mathrm{H}_{2} \mathrm{O}$ Considering 1 mole of biomass $\mathrm{BOD}_{4}=160 \mathrm{gm} / \mathrm{mol}$ $\therefore \quad \mathrm{BOD}_{5}=\mathrm{BOD}_{4}\left(1-\mathrm{e}^{-\mathrm{k}_{0} \mathrm{t}}\right)$ $=160\left(1-e^{-0.23 \times 5}\right)$ $=160 \times 0.6833$ $=109.34 \mathrm{gm} / \mathrm{mol}$ Calculation of TOC: $\mathrm{TOC}=\frac{12 \times 5}{113} \times 113=60 \mathrm{gm} / \mathrm{mol}$ ( $\because 1$ mole biomass is considered) $\therefore$ Required Ratio $=\frac{109.34}{60}=1.82$

Concentration of $SO_2$ in ppm = 0.04 Let's equivalent concentration in $=\mu g/m^3$ is $x$ . $x\mu g$ of $SO_2$ present in $1m^3$ of air. We know, 1 mole of $SO_2$ (at $0^{\circ}C$ , 1 atm) has volume of 22.4 L. at $25^{\circ}C$ and 1 atm, volume of $SO_2$ $=\frac{22.4}{273+0}\times (273+25)=24.45 lit$ $\frac{x \times 10^{-6}}{64}$ mole of $SO_2$ has volume $\frac{24.45 \times x \times 10^{-6}}{64} lit \; \in \; 1m^3$ of air. $0.382x \times 10^{-3}m^3$ of $SO_2$ present in $10^{6}m^3$ of air. $0.382x \times 10^{-3}=0.04$ $x=104.7 \approx105$ 0.04 ppm of $SO_2=105\mu g/m^3$ of $SO_2$

VOC emission = 5 Kg/h Capturing Efficiency =95% Fugitive emission = (5/100)*5 Kg/h=0.25 Kg/h

Indoor environment photochemical smog does not form.

$\text{Concentration} \propto \frac{1}{\text{wind speed}}$

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

Given, discharge to be passed = $12 \mathrm{~m}^{3} / \mathrm{s}$ Out of 5 sections, 4 will operate at a time. Each bag is Thus, surface area of each bag $=\pi \times 0.32 \times 5 \mathrm{~m}^{2}=5.0265 \mathrm{~m}^{2}$ Total surface area required wr.t. given discharge $=\frac{12 \mathrm{~m}^{3} / \mathrm{s}}{6 \mathrm{~m}^{3} / \mathrm{min}-\mathrm{m}^{2}}=\frac{12 \mathrm{~m}^{3} / \mathrm{s}}{\frac{6}{60} \mathrm{~m}^{3} / \mathrm{s}-\mathrm{m}^{2}}=120 \mathrm{~m}^{2}$ $\therefore$ Working bag filters required $=\frac{120 m^{2}}{5.0265 m^{2}}=$ 23.87 or 24 bag filters But, since it is asked total (which includes a standby set also), then,Total no. of bag filters $=24 \times \frac{5}{4}=30$

Pyrolysis is an endothermic process as there is a substantial heat input required to raise the biomass to the reaction temperature.

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 $\mu m$ , but in actual they are not practical for removal of particles much less than 50 $\mu m$ in size. Hence, gravitational settling chamber is most suitable choice for removal of suspended particles of size 50 $\mu m$ . After the removal of 50 $\mu m$ size particles, to remove 2 $\mu m$ 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 $\mu m$ in diameter.

$41.6\mu$ mole of gas volume of $10^{-6} m^3$

$\underbrace{C}_{12}\underbrace{O}_{2}\rightarrow CO_2$ Oxygen required for 350 tonne/day $\frac{32}{12}\times 350=933.33$ $\underbrace{4H}_{4}+\underbrace{O_2}_{32}\rightarrow 2H_2O$ Oxygen required for 100 tonne/day $\frac{32}{4}\times 100=800$ $\underbrace{S}_{32}+\underbrace{O_2}_{32}\rightarrow SO_2$ Oxygen required for 60 tonne/day $\frac{32}{32}\times 60=60$ $\underbrace{N}_{14}+\underbrace{O}_{32}\rightarrow NO_2$ Oxygen required for 30 tonne/day $\frac{32}{14}\times 30=68.57$ Total $O_2$ = 1861.9 tonne/day Available $O_2$ in waste = 260 tonne/day Required =1861.9-260=1601.9 tinne/day Amount of air required $=\frac{1601.9}{0.23}=6964.78 \; \text{tonne/day} \simeq 6965 \; \text{tonne/day}$

Hydrocarbons and nitrogen oxides are considered primary air pollutants.