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Test: Environmental Engineering- 5 - Civil Engineering (CE) MCQ


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10 Questions MCQ Test GATE Civil Engineering (CE) 2025 Mock Test Series - Test: Environmental Engineering- 5

Test: Environmental Engineering- 5 for Civil Engineering (CE) 2024 is part of GATE Civil Engineering (CE) 2025 Mock Test Series preparation. The Test: Environmental Engineering- 5 questions and answers have been prepared according to the Civil Engineering (CE) exam syllabus.The Test: Environmental Engineering- 5 MCQs are made for Civil Engineering (CE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Environmental Engineering- 5 below.
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Test: Environmental Engineering- 5 - Question 1

Which of the following biological unit process is based on a suspended growth system

1. Activated sludge process

2. Aerated lagoon

3. Rotating Biological contractors

4. Aerobic and Anaerobic Digester

Detailed Solution for Test: Environmental Engineering- 5 - Question 1

Test: Environmental Engineering- 5 - Question 2

The maximum efficiency of BOD removal is achieved in which of the following?

Detailed Solution for Test: Environmental Engineering- 5 - Question 2

The main advantage of oxidation ditch is the ability to achieve removal performance objective with low operational requirements and eminence costs.An oxidation ditch is a modified activated sludge biological treatment process that uses long solids retention times (SRTs) to remove biodegradable organics. The typical oxidation ditch is equipped with aeration rotors or brushes that provide aeration and circulation. The wastewater moves through the ditch at 1 to 2ft/s. The ditch may be designed for continuous or intermittent operation. Because of this feature, this process may be adaptable to the fluctuations in flows and loadings associated with recreation area wastewater production. Several manufacturers have developed modifications to the oxidation ditch design to remove nutrients in conditions cycled or phased between the anoxic and aerobic states.

Oxidation ponds, also called lagoons or stabilization ponds, are large, shallow ponds designed to treat wastewater through the interaction of sunlight bacteria, and algae. Algae grow using energy from the sun and carbon dioxide and inorganic compounds released by bacteria in water. During the process of photosynthesis, the algae release oxygen needed by aerobic bacteria. Mechanical aerators are sometimes installed to supply yet more oxygen, thereby reducing the required size of the pond. Sludge deposits in the pond must eventually be removed by dredging. Algae remaining in the pond effluent can be removed by filtration or by a combination of chemical treatment and settling.

Test: Environmental Engineering- 5 - Question 3

In aerobic environment, nitrosomonas convert

 

Detailed Solution for Test: Environmental Engineering- 5 - Question 3

Test: Environmental Engineering- 5 - Question 4

The main constituents of gas generated during the anaerobic digestion of sewage sludge are

Detailed Solution for Test: Environmental Engineering- 5 - Question 4

Anaerobic digestion is normally adopted for primary sludge because primary sludge contains large amount of readily available organics that would induce rapid growth, of biomass if treated anaerobically.

Anaerobic decomposition process produce less biomass and primary function of anaerobic digester is to convert as much of sludge as possible to end products like liquid and gases while producing little biomass.

Waste-water contains wide varieties of organism.

These organisms are broadly classified as

a) Acid formers

b) Methane formers

Acid formers: Acid formers consists of facultative and anaerobic bacteria and include organisms which solubilize the organic acid through hydrolysis.The soluble end products are then fermented to acids and alcohol of lower molecular weight.

Methane formers: Methane formers are strictly anaerobic and converts acid and alcohol alongwith hydrogen and CO2 to methane.

*Answer can only contain numeric values
Test: Environmental Engineering- 5 - Question 5

What would be the concentration of returned sludge (in mg/l) if the sludge volume index for an activated sludge is 92 ml per gm


Detailed Solution for Test: Environmental Engineering- 5 - Question 5

Concentration of returned sludge =

= 10869.56 mg/l

*Answer can only contain numeric values
Test: Environmental Engineering- 5 - Question 6

A conventional activated sludge treatment plant has to treat wastewater of 2250000 litres/day, having a five day BOD of 225 mg/l. The mixed liquor suspended solids concentration in the aeration tank is 2250 mg/l and concentration of biomass leaving the system is 22.5 g/m3. If the volume of the aeration tank is 225 m3, then the average time (in days) for which the biomass stays in the system will be


Detailed Solution for Test: Environmental Engineering- 5 - Question 6

The average time for which the biomass stays in the system is called sludge age

where V = volume of the tank

X = mixed liquor suspended solids concentration in the aeration tank

Qw = flow of wastewater

Xu = Biomass leaving the aeration tank per day

Test: Environmental Engineering- 5 - Question 7

A sedimentation tank is treating 2 MLD of sewage per day containing 225 mg/L of suspended solids. The tank removes 30% of suspended solids. The moisture content of sludge is 98%. If the sludge is reduced to half of its volume, then the moisture content of the reduced volume sludge is –

Detailed Solution for Test: Environmental Engineering- 5 - Question 7


 

⇒ 100 – P = 2 × 2

⇒ P = 100 – 4

⇒ P = 96%

Test: Environmental Engineering- 5 - Question 8

The efficiency of a 50 m diameter and 1.5 m deep single stage high rate trickling filter based upon the data given below is _________.

Given Data:

Sewage flow = 10 MLD

Recirculation factor = 1.25

BOD of raw sewage = 200 mg/L

BOD removed in primary clarifier = 35%

Detailed Solution for Test: Environmental Engineering- 5 - Question 8

The efficiency of a high rate single stage trickling filter is given as:

Where,

Y = Total organic loading in kg/day

V= Filter volume in hectare. m

F = Recirculation factor = 1.25 (given)

BOD to be removed in filter

= Total BOD of sewage – BOD removed in primary clarifier

= 200 mg/L – 35% of 200 mg/L

= 200 – 70

= 130 mg/L

Y = total organic loading

= 130 mg/L × 10 × 106 L/d

= 1300 kg/d

n = 79.26%

Test: Environmental Engineering- 5 - Question 9

For a colony of 10000 persons having sewage flow rate 250 lpcd with a BOD of 320 mg/l and organic loading of 200 Kg/day/ha, the area of the oxidation pond required for treating the sewage of the colony is _______

Detailed Solution for Test: Environmental Engineering- 5 - Question 9

Total BOD produced per day = 
Organic loading rate = 200 kg/day/ha

*Answer can only contain numeric values
Test: Environmental Engineering- 5 - Question 10

The activated sludge process schematic flow diagram is shown in the figure below:

where, Q, Qw = flow rate, m3/d

θc = sludge age, days

X, XR = microorganism- concentration (mixed – liquor volatile suspended solids or MLVSS)

The volume of sludge that must be wasted each day if wastage is accomplished from point A will be _________m3/day.


Detailed Solution for Test: Environmental Engineering- 5 - Question 10


According to definition

sludge age 

Qw = 100 m3/day

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