Depending on these conditions different types of plume behavior are shown where X-axis indicates the variation of wind velocity, temperature, horizontal movement of plume while Y-axis indicates the variation with an increase in height, vertical movement of plume.

Concept:

1. Electrostatic Precipitators:-

• In electrostatic precipitators, the emitted gas flue gas is passed through a highly ionized atmosphere high-voltage field; and in that zone particulates get electrically charged and get separated from the gaseous stream with the help of electrostatic forces. Fine particles up to 1 μm wet or dry can be removed easily in an electrostatic pressure.
• Four basic steps required in the operation of a high-voltage single-stage electrostatic precipitator:

a) electrical charging of the particulates

b) collection of charged particles on a grounded surface

c) neutralization of the charge at the collector

d) removal of the particulate for disposal.

2. Gravitational Settling Chambers:

• Settling chambers in air-pollution control systems provide enlarged areas to minimize horizontal velocities and allow time for the vertical velocity to carry the particle to the floor.
• The emitted smokes, when made to pass through a settling chamber, drop some of their larger sized particles in the chamber as per Stoke’s Law. The largest size particle (d) that can be removed with 100% efficiency in such a chamber of length L and height H is given by

​\(d=C.\sqrt{\frac{18\mu .{{v}_{h}}.H}{g.L.{{\rho }_{p}}}}\)

Where, vh = Horizontal velocity of gas passing through the chamber

• Simple to design and maintain, and low-pressure flow.
• It requires larger space for installation and has low collection efficiency for small-sized particles.
• Only larger sized particles are separated.
• Although theoretically they should be able to remove particulates down to 5 or 10 μm, in actual, they are not practical for the removal of particles much less than 50 μm in size.

3. Fabric Filters

• In fabric filters, the gas stream laden with particles is passed through a woven or felted fabric that filters out particulate matter, allowing the gaseous matter to flow on.
• Small, particles are initially retained on the fabric by direct interception, inertial impaction, diffusion, electrostatic attraction, and gravitational settling.
• The collection of sub-micron particles is accomplished by sieving after a mat of dust gets found on the fabric.
• Filter bags, are capable of removing most particles as small as 0.5 μm and will remove substantial quantities of particles as small as 0.1 μm with an efficiency greater than 99%.

4. Wet Scrubbers:

• Wet scrubbers remove the particulates from the incoming gaseous stream by allowing the flue gases to flow up against a falling water stream.  
• When aqueous chemical solutions, other than water, like lime, oil etc. it is used for the removal of gaseous pollutants also from the flue gases.  
• Generally, collectors operating at very low pressure loss remove only medium- to coarse-size particles, while collectors operating at higher pressure losses are highly efficient at removing fine particles.
• Most commonly used wet collectors are— the spray tower, the wet cyclone scrubber, and the venturi scrubber.​

Concept:

Catalytic Combustion: It is used when combustible materials in the waste gas are too low to make direct-flame incineration feasible. A catalyst accelerates the rate of oxidation without itself undergoing a chemical change, thus reducing the residence time required for incineration.

Examples of Catalysts:

(i) Vanadium pentoxide for removing SO₂

(ii) Platinum metals – for treating NO₃

(iii) Activated alumina – for hydrocarbons

(iv) Palladium II (Pd II), and Cu (II) – to oxidise CO and CO₂.

According to Noise pollution (Regulation and control) Rules 2000 of India, the day and night time noise limit for various zones is categories into 4 group as mentioned in the table below

Concept:

Sound power level from distance R₁ = L₁

Sound power level from distance R₂ = L₁

\(\rm{\therefore {L_2} = {L_1} - \left| {20\log \left( {\frac{{{R_2}}}{{{R_1}}}} \right)} \right|}\)

Calculation:-

L₁ = 50 dB, R₁ = 100 m, L₂ = 55 dB, R₂ = ?

\(\rm{\therefore {L_2} = 55\;dB = 50\;dB - \left| {20\log \left( {\frac{{{R_2}}}{{100}}} \right)} \right|}\)

\(\rm{\Rightarrow 55 - 50 = - \left| {20||\log \left( {\frac{{{R_2}}}{{100}}} \right)} \right|}\)

\(\rm{\Rightarrow - \left| {\log \left( {\frac{{{R_2}}}{{100}}} \right)} \right| = \frac{5}{{20}}}\)

\(\rm{\frac{{{R_2}}}{{100}} = {10^{ - 5/20}}}\)

\(\rm{\frac{{{R_2}}}{{100}} = 0.5623}\)

Concept:

Adiabatic Lapse Rate

• When a parcel of air which is hotter and lighter than the surrounding air is released, then naturally it tends to rise up until it reaches a level at which its own temperature and density becomes equal to that of the surrounding air. This rate of decrease of temperature with height is called adiabatic lapse rate.
• Dry air expanding and cooling adiabatically cools at rate of 9.8°C per km and it is called dry adiabatic lapse rate. In saturated (wet) air, this rate is calculated to be 6°C per km, and is known as wet adiabatic lapse rate

Calculation:

Saturated air cools at a rate of 6°C per km.

Total distance ascended by the air parcel = 900 m = 0.9 km

Initial temperature = 60°C

Temperature drop = 6°C per km = 0.9 × 6 = 5.4°C

∴ Final Temperature = 60 – 5.4 = 54.6°C   

Important Point:

Wet adiabatic lapse rate is smaller than the dry adiabatic lapse rate due to release of latent heat or condensation of water vapour within the saturated parcel of rising air.

Concept:

L_eq Concept:

L_eq is defined as the constant noise level, which over a given time, expands the same amount of energy as is expanded by the fluctuating levels over the same time. This value is expressed as

\({L_{eq}} = 10\log \mathop \sum \limits_{i = 1}^{i = n} {\left( {10} \right)^{\frac{{{L_1}}}{{10}}}} \times {t_i}\)

Where,

n = Total number of sound samples

L₁ = The noise level of any ith sample

t_i = Time duration of i^th sample, expressed as fraction of total sample time.

Calculation:

Total sample time (T) = 75 minutes

\({L_{eq}} = 10\log \mathop \sum \limits_{i = 1}^{i = n} {\left( {10} \right)^{\frac{{{L_i}}}{{10}}}} \times {t_i}\)

\(= 10{\log _{10}}\left[ {\left( {{{10}^{\frac{{{L_1}}}{{10}}}} \times \frac{{{t_1}}}{T}} \right) + \left( {{{10}^{\frac{{{L_2}}}{{10}}}} \times \frac{{{t_2}}}{T}} \right) + \left( {{{10}^{\frac{{{L_3}}}{{10}}}} + \frac{{{t_3}}}{T}} \right)} \right]\)

\(= 10{\log _{10}}\left[ {\left( {{{10}^{\frac{{60}}{{10}}}} \times \frac{{30}}{{75}}} \right) + \left( {{{10}^{\frac{{90}}{{10}}}} \times \frac{{15}}{{75}}} \right) + \left( {{{10}^{\frac{{75}}{{10}}}} \times \frac{{30}}{{75}}} \right)} \right]\) = 83.28 dB

Calculation:

E₁ = 90%, i.e. the first precipitator removes 90% of the particulates.

The second precipitator removes E₂ % of the remaining 10%.

\(\therefore {\eta _{overall}} = {E_1} + \frac{{10}}{{100}} \times {E_2}\)

\(\Rightarrow 95 = 90 + \frac{{10}}{{100}} \times {E_2}\) ⇒ E₂ = 50%

Concept:

Conversion of air pollutants concentration:

\({\rm{x\;}}\left( {{\rm{\mu g}}/{{\rm{m}}^3}} \right) = {\rm{x\;}}\left( {{\rm{ppb}}} \right) \times \frac{{{{\rm{M}}_{\rm{x}}}}}{{{\rm{V\;}}\left( {{\rm{L}}/{\rm{mole}}} \right){\rm{\;}}}}\)

\({\rm{x\;}}\left( {{\rm{\mu g}}/{{\rm{m}}^3}} \right) = {\rm{x\;}}\left( {{\rm{ppm}}} \right) = \frac{{{{\rm{M}}_{\rm{x}}}}}{{{\rm{V\;}}\left( {{\rm{L}}/{\rm{Mole}}} \right)}} \times {10^3}\)

P × V = n × R × T

At another condition

\(\frac{{{{\rm{P}}_1}{{\rm{V}}_1}}}{{{{\rm{T}}_1}}} = \frac{{{{\rm{P}}_2}{{\rm{V}}_2}}}{{{{\rm{T}}_2}}}\)

Also, \({\rm{V}} = 0.082 \times \frac{{{\rm{T}}\left( {{\rm{Kelvin}}} \right)}}{{{\rm{P\;}}\left( {{\rm{atm}}} \right)}}\)

Calculation:

PV = nRT

\(\therefore {\rm{v}} = \frac{{{\rm{RT}}}}{{\rm{P}}} = \frac{1}{{{\rm{P}}/{\rm{RT}}}} = \frac{1}{{41.6{\rm{\;}}\left( {{\rm{mol}}/{{\rm{m}}^3}} \right)}} = \frac{{1000}}{{41.6}} = 24.03{\rm{\;l}}/{\rm{mole}}\)

\(\therefore {\rm{x}} = 30 \times \frac{{46}}{{24.03}} = 57.408{\rm{\;\mu g}}/{{\rm{m}}^3}\)

Given: Total houses = 3000, Average person in each household = 4, Solid waste generation rate = 500 g/capita/day, Solid waste goes to landfill = 85%, Weight of solid waste which goes to landfill = 3000 × 4 × 0.5 × 0.85 × 20 × 365 = 37.23 × 10⁶ kg, and Density of uncompacted solid waste = 180 kg/m³

Calculation:

Volume of uncompacted solid waste = \(\frac{{37.23 \times {{10}^6}}}{{180}}\) = 206833.33 m³

It is compacted to 60% of initial volume,

Volume of compacted solid waste = 124100 m³, Landfill cover volume = 0.05 × 124100 = 6205 m³