Environmental Engineering Test 5

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

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