Environmental Engineering MCQ Quiz - Objective Question with Answer for Environmental Engineering - Download Free PDF

Concept:

Biological Oxygen Demand(BOD):

Biological Oxygen Demand or Biochemical oxygen demand (BOD) is the amount of dissolved oxygen used by microorganisms to break down organic matter in water.

A low BOD is an indicator of good quality water, while a high BOD indicates polluted water.

BOD = ( DOi - DFf ) × Dilution factor

Where,

BOD = Biochemical oxygen demand in ppm or mg/lit

DOi = Initial dissolved oxygen in mg/lit.

DOf =Final dissolved oxygen in mg/lit.

Dilution factor \(= \frac{{Volume\ of\ the\ diluted\ sample}}{{{\rm{Volume \ of\ the\ undiluted\ sewage\ sample}}}} \)

Calculation:

Given:

Dilution factor \(= \frac{{100}}{{{\rm{\% \;solution}}}} = \frac{{100}}{1} = 100\)

Depletion of oxygen = 3 ppm

BOD5 = Depletion of oxygen × Dilution factor

BOD5 = 3 × 100 = 300 ppm

Concept:

Reservoir Yield:

• It means the quantity of water which can be withdrawn from storage in the reservoir. Reservoir yield is determined by the rate of flow of the stream into the reservoir, losses due to evaporation from the reservoir surface, and the volume of water impounded in the reservoir 
• It is the amount of water that can be supplied by the reservoir in a specified interval of time.
• The specified time interval may vary from a day for a small distribution reservoir to a month or year for large conservation reservoirs.
• If we say that three million cubic meters of water can be supplied from a reservoir in a year then its yield is 3000000 m3/year.
•  The yield of the reservoir is dependent upon the inflow and thus varies from time to time.

Firm yield.

•  It is also known as Safe yield.
•  It is the maximum quantity of water that can be supplied from the reservoir with full guarantee during the worst dry period.

 Design Yield.

• The critical period for a reservoir is generally considered when natural flow in the reservoir is minimum.
• Hence a lower value than the guaranteed yield or safe yield may be taken for design purposes.
• This yield whose value is smaller than the safe or firm yield is known as design yield.
•  The value of design yield for a reservoir to be used for water supply is taken less than the safe yield.
•  In the case of reservoirs used for irrigation purpose, the design yield may be taken slightly more than the safe yield as crops can tolerate some deficiency of water during the exceptionally dry season

A Secondary Yield.

• The quantity of water available more than safe yield is known as secondary yield.
• This yield is available during a period of high inflows.
• This secondary yield of the reservoir can be used either to generate extra hydroelectric power or for irrigation of extra lands.

Average Yield.

•  The arithmetic average of the safe yield and the secondary yield considered for a number of years is known as average yield.
•  The storage capacity of the reservoir and its yield are very much interdependent.
• The water is stored in the reservoir to fulfill the safe yield requirements.
• If the capacity of the reservoir is more it can certainly provide more water and hence yield is more.
• The reservoirs are designed to meet specific water demands.

Explanation:

Oxygen sag curve of river manifest dissolved oxygen deficit.

The difference between saturated dissolved oxygen content and the actual dissolved oxygen content in the stream at any point during self-purification process is called Oxygen sag 

Oxygen sag curve or oxygen deficit curve is obtained by algebraic addition of deoxygenation and reoxygenation curve.

Concept:

The settling velocity of sand particles is given by Stoke's Law,

\({{\rm{V}}_{\rm{s}}} = \frac{{g \times \left( {G - 1} \right) \times {d^2}}}{{18\;ν }}\)...... (1)

G -Specific gravity of sand particles

ν - Kinematic viscosity

d - Diameter of sand particles

Calculation:

Given:

Let V₁ and V₂ be settling velocities of larger and smaller grain respectively

V₁ = 0.5 mm/sec

Diameter of larger grain d₁ = 0.075 mm

Diameter of smaller grain d₂ = 0.0075 mm

g is constant, specific gravity identical for both grains, both the particles are falling through water hence ν remains same. Hence equation 1 becomes

⇒ \(\frac{{{V_1}}}{{{V_2}}}\; = \;{\left( {\frac{{{d_1}}}{{{d_2}}}} \right)^2}\)

⇒ \(\frac{{{0.5}}}{{{V_2}}}\; = \;{\left( {\frac{{{0.075}}}{{{0.0075}}}} \right)^2}\)

⇒ V₂ = 0.005 mm/sec

Explanation:

Secondary pollutant:

• The primary pollutants often react with one another or with water vapor, aided and abetted by the sunlight, to form entirely a new set of pollutants, called the secondary pollutants. 
• These are the chemical substances, which are produced from the chemical reactions of natural or anthropogenic pollutants or due to their oxidation, etc., caused by the energy of the sun.

Examples:

1) Sulphuric acid (H2SO4)

2) Formaldehydes

3) Photochemical smog

4) Ozone

5) Peroxy Acetyl – nitrate (PAN)

Primary pollutant: 

• Pollutants are emitted directly from identifiable sources, either from natural hazardous events like dust storms, volcanoes, etc, or from human activities like burning of wood, coal, oil in homes or industries or automobiles, etc.

Examples:

1) SO2

2) CO & CO2

3) NO, NO2

4) Volatile organic compound

5) Suspended Particulate Matter (SPM)

6) Radioactive compound

7) Halogen compound

The correct answer is Filtration.

Key Points

Filtration

• Filtration is a process used to separate solids from liquids or gases by using a filter medium that allows fluid to pass through but not solid. 
• Filtration can be mechanical-biological or physical. 
• It does not help in disinfecting water. 
• If you want to disinfect the water you can boil it or you can use alum for chlorine tablets. 

Boiling

• Boiling is used to kill pathogenic bacteria, viruses, and protozoa. 

Chlorination

• Chlorine kills a large variety of microbial waterborne pathogens.
• Chlorination is the process of adding chlorine to the water to disinfect it and kill pathogens. 
• Amount of chlorine required for water disinfection around 1-16 milligrams per litre of water.

Coagulation

• Alum acts as an electrolyte that helps in settling the suspended matter in water.
• The process of adding alum to disinfect water is called coagulation. 

Concept:

• Water softening: It is the process of hardness removal from the water. It is caused by multivalent cation and affects water quality.
   
• Lime soda method: It is a water softening method in which lime and soda ash are added to the water, which causes the precipitation of multivalent cation as CaCO3.
• Precipitation of CaCO3 occurs only when the pH of water is greater than 9, so in case of less pH alkalinity is added to the water. In this process small amount of Ca2+ and Mg2+ precipitates very late, which will create incrustation in the pipe, so to avoid this recarbonation is done to dissolve back this small amount of cation.
• Due to this, the method does not give zero hardness.
   
• Ion Exchange Process: Ion-exchange resin, (zeolite) exchanges one ion from the water being treated for another ion that is in the resin (sodium is one component of softening salt, with chlorine being the other). Zeolite resin exchanges sodium for calcium and magnesium. It can produce water with zero hardness.

Explanation:

The odour characteristic is used for classifying odour of a given water sample as per IS 3025 (part-5) 1983 as follows:

1. Degrees of sweetness

2. Degrees of pungency

3. Degrees of smokiness

4. Degrees of Rottenness

Concept:

Odour and taste are expressed by threshold odour number (T.O.N.) as it represents the dilution ratio at which odour can not be detected.

\(T.O.N. = \frac{{A + B}}{A} = \frac{{volume\ of\ diluted\ sample}}{{Volume\ of\ undiluted\ sample}}\)

where A = Volume of water sample undiluted 

B = volume of distilled water required to be added to remove the odour.

Calculation:

Given,

A + B = 250 ml

A = 25 ml

\(T.O.N. = \frac{{A + B}}{A} = \frac{{250}}{{25}}\)

T.O.N. = 10

Mistake Points

• Generally in these question volume of water to be added (B), is given.
• But In the question, it is given " sample was diluted to ", which means that the final volume of the sample after dilution(A+B) is given 

Concept:

The peak flow can be considered as 1.5 times the annual average daily flow.

For a design of the treatment facility, the peak factor is considered as 1.5 times the annual average daily flow.

The minimum flow passing through sewers is also important to develop self-cleansing velocity to avoid silting in sewers. This flow will generate in the sewers during late night hours. The effect of this flow is more pronounced on lateral sewers than the main sewers.

Sewers must be checked for minimum velocity as follows:

Minimum daily flow = 2/3 Annual average daily flow

Minimum hourly flow = 1/2 Minimum daily flow

Minimum hourly flow = 1/3 Annual average daily flow

Followings are the classification of secondary treatment units:

Explanation:

Activated sludge process:

• The essential features of the activated sludge process are an aeration stage, solids-liquid separation following aeration, and a sludge recycle system.
• Wastewater after primary treatment enters an aeration tank where the organic matter is brought into intimate contact with the sludge from the secondary clarifier.
• It requires less space, does not produce obnoxious odor, and requires less time for wastewater treatment.
• It requires skilled supervision

Followings are the classification of secondary treatment units:

As per IS 10500: 2012, the following are the various permissible limits for different parameters:

Fluoride or fluorine deficiency is a disorder that may cause increased dental caries (or tooth decay) is the breakdown of dental tissues by the acidic products released by the bacterial fermentation of dietary carbohydrates.

Excess of nitrates is harmful to infants and causes Methemoglobinemia or Blue baby disease.

Concept:

The population after n^th decade by using Arithmetic mean Method is given as:

\({P_n} = P + N\bar X\)

Where

P is the Present Population

N is no of decades of which population is to be calculated

X̅ is the average increase in population.

Calculation

\(\bar X = \frac{{9,000\; + \;7,000\; + \;12,000}}{3} = 9333.333\)

The population at end of 2011 i.e. after 3 decodes from 1981.

\({P_{{g_{011}}}} = 1,28,000 + 3 \times 9333.333\)

\({P_{{g_{011}}}} = 1,56,000\)

Explanation:

Detention time for different type of Treatment unit