Environmental Engineering

Question 1
A system of seven river segments is shown in the schematic diagram. The \mathrm{R}_{1}{ }^{\prime} \mathrm{s}, \mathrm{Q}_{1}{ }^{\prime} \mathrm{s}, and \mathrm{C}_{\mathrm{i}} 's ( \mathrm{i}=1 to 7 ) are the river segments, their corresponding flow rates, and concentrations of a conservative pollutant, respectively. Assume complete mixing at the intersections, no additional water loss or gain in the system and steady state condition. Given : \mathrm{Q}_{1}=5 \mathrm{~m}^{3} / \mathrm{s} ; \mathrm{Q}_{2}=15 \mathrm{~m}^{3} / \mathrm{s} ; \mathrm{Q}_{4}=3 \mathrm{~m}^{3} / \mathrm{s} ; \mathrm{Q}_{6}=8 \mathrm{~m}^{3} / \mathrm{s} ; C_{1}=8 \mathrm{~kg} / \mathrm{m}^{3} ; C_{2}=12 \mathrm{~kg} / \mathrm{m}^{3} ; C_{6}=10 \mathrm{~kg} / \mathrm{m}^{3}. What is the steady state concentration in \mathrm{kg} / \mathrm{m}^{3}, rounded off to two decimal place) of the pollutant in the river segment 7 ? ____

GATE CE 2023 SET-2      Design of Sewers and Sewerage System
Question 1 Explanation: 

At 1-1, for \mathrm{R}_{3}
\begin{aligned} \mathrm{Q}_{3} & =\mathrm{Q}_{1}+\mathrm{Q}_{2} \\ & =5+15=20 \mathrm{~m}^{2} / \mathrm{sec} \\ \mathrm{C}_{3} & =\frac{\mathrm{Q}_{1} \mathrm{C}_{1}+\mathrm{Q}_{2} \mathrm{C}_{2}}{\mathrm{Q}_{1}+\mathrm{Q}_{2}} \\ &=\frac{5 \times 8+15 \times 12}{5+15} \\ & =11 \mathrm{Kg} / \mathrm{m}^{3} \end{aligned}

At 2-2, For \mathrm{R}_{5}
\mathrm{Q}_{5}=\mathrm{Q}_{3} \mathrm{Q}_{4}=10-3=17 \mathrm{~m}^{3} / \mathrm{sec}
\mathrm{C}_{5}=\mathrm{C}_{3}=11 \mathrm{Kg} / \mathrm{m}^{3}

At 3-3 for \mathrm{R}_{7}
\begin{aligned} \mathrm{Q}_{7} & =\mathrm{Q}_{6}+\mathrm{Q}_{5}=17+8=25 \mathrm{~m}^{3} / \mathrm{sec} \\ \mathrm{C}_{7} & =\frac{\mathrm{Q}_{6} \mathrm{C}_{6}+\mathrm{Q}_{6} \mathrm{C}_{5}}{\mathrm{Q}_{6}+\mathrm{Q}_{5}} \\ & =\frac{8 \times 10+17 \times 11}{8+17}=10.68 \mathrm{Kg} / \mathrm{m}^{3} \end{aligned}
Answer : 10.6 \mathrm{~kg} / \mathrm{m}^{3}.
Question 2
The theoretical aerobic oxidation of biomass (\mathrm{C} 5 \mathrm{H} 7 \mathrm{O} 2 \mathrm{~N} ) is given below:

\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}

The biochemical oxidation of biomass is assumed as a first-order reaction with a rate constant of 0.23 / \mathrm{d} at 20^{\circ} \mathrm{C} (logarithm to base e). Neglecting the second-stage oxygen demand from its biochemical oxidation, the ratio of \mathrm{BOD} 5 at 20^{\circ} \mathrm{C} to total organic carbon (TOC) of biomass is (round off to two decimal places).
[Consider the atomic weights of \mathrm{C}, \mathrm{H}, \mathrm{O} and \mathrm{N} as 12 \mathrm{~g} / \mathrm{mol}, 1 \mathrm{~g} / \mathrm{mol}, 16 \mathrm{~g} / \mathrm{mol} and 14 \mathrm{~g} / \mathrm{mol}, respectively]
GATE CE 2023 SET-2      Air and Noise Pollution
Question 2 Explanation: 
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

Question 3
Which of the following statements is/are TRUE for the aerobic composting of sewage sludge?
Bulking agent is added during the composting process to reduce the porosity of the solid mixture
Leachate can be generated during composting
Actinomycetes are involved in the process
In-vessel composting systems cannot be operated in the plug-flow mode
GATE CE 2023 SET-2      Disposal of Sewage Effluents
Question 3 Explanation: 
A bulking agent (wood ships shredded leaves or types) is intended to support the structure of sludge by increasing its porosity to encourage effective aeration.

In-vessel composting systems can be divided into plug flow and agitated bed modes. In plug flow mode, the relationship between particles in the composting mass stays the same through the process and the system operates on first-in, first-out principle. In an agitated bed system, composting material is mixed mechanically during the processing.

The heat produced in aquatic composting cause any surplus moisture to be extracted as water vapour and hence no leachate is produced.

Actinomycetes are aerobic, spore forming positive bacteria. These are thermophilic species and their development depends on aerobic conditions, temperature and water content. The heat produced in aerobic composting helps support the growth of these species.

Leachate can generated from vegetable wastes.
Question 4
In the context of water and wastewater treatments, the correct statements are:
particulate matter may shield microorganisms during disinfection
ammonia decreases chlorine demand
phosphorous stimulates algal and aquatic growth
calcium and magnesium increase hardness and total dissolved solids
GATE CE 2023 SET-2      Treatment of Waste Water
Question 4 Explanation: 
Total dissolved solids comprise of inorganic salts, principally calcium, magnesium, potassium, sodium bicarbonates, chlorides, sulfates and some small amount of organic matter that are dissolved in water.
Hardness is due to presence of multivalent cations. In most cases, it is due to the presence of soluble bicarbonates, chlorides and sulfates of calcium and magnesium.
Too much phosphorous can cause increased growth of algae and large aquatic plants. It can result in decreased levels of dissolved oxygen and the process is called 'Eutrophication'.
Ammonia creates a large chlorine demand, as the ammonia is converted to nitrogen gas and the chlorine becomes chloride during disinfection of water by chlorination.
Presence of particulate matter decreases the effectiveness of UV disinfection or chlorine disinfection by shielding the targeted microorganisms.
Question 5
Match the following air pollutants with the most appropriate adverse health effects:
\begin{array}{|l|l|} \hline \textbf{Air pollutant} & \textbf{Health effect to human and/or test animal} \\ \hline \text{(P) Aromatic hydrocarbons} & \text{(I) Reduce the capability of the blood to carry oxygen} \\ \hline \text{(Q) Carbon monoxide} & \text{(II) Bronchitis and pulmonary emphysema}\\ \hline \text{(R) Sulfur oxides}& \text{(III) Damage of chromosomes} \\ \hline \text{(S) Ozone} & \text{(IV) Carcinogenic effect} \\ \hline \end{array}
(P) - (II), (Q) - (I), (R) - (IV), (S) - (III)
(P) - (IV), (Q) - (I), (R) - (III), (S) - (II)
(P) - (III), (Q) - (I), (R) - (II), (S) - (IV)
(P) - (IV), (Q) - (I), (R) - (II), (S) - (III)
GATE CE 2023 SET-2      Air and Noise Pollution
Question 5 Explanation: 
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.

There are 5 questions to complete.

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